Graduate Texts
in Mathematics
Bruce E. Sagani
P The Symmetric
p Group
Representations,
Combinatorial
Algorithms, and
Symmetric Functions
Second Edition
Graduate Texts in Mathematics
203
Editorial Board
S. Axler F.W. Gehring K.A. Ribet
Springer Science+Business Media, LLC
Graduate Texts in Mathematics
1 Takeuti/Zaring. Introduction to
Axiomatic Set Theory. 2nd ed.
2 OxTOBY. Measure and Category. 2nd ed.
3 Schaefer. Topological Vector Spaces.
2nd ed.
4 Hilton/Stammbach. A Course in
Homological Algebra. 2nd ed.
5 Mac Lane. Categories for the Working
Mathematician. 2nd ed.
6 Hughes/Piper. Projective Planes.
7 Serre. a Course in Arithmetic.
8 Takeuti/Zaring. Axiomatic Set Theory.
9 Humphreys. Introduction to Lie Algebras
and Representation Theory.
1 0 Cohen. A Course in Simple Homotopy
Theory.
1 1 Conway. Functions of One Complex
Variable I. 2nd ed.
12 Beals. Advanced Mathematical Analysis.
13 Anderson/Fuller. Rings and Categories
of Modules. 2nd ed.
14 Golubitsky/Guillemin. Stable Mappings
and Their Singularities.
15 Berberian. Lectures in Functional
Analysis and Operator Theory.
16 Winter. The Structure of Fields.
17 Rosenblatt. Random Processes. 2nd ed.
1 8 Halmos. Measure Theory.
19 Halmos. A Hilbert Space Problem Book.
2nd ed.
20 Husemoller. Fibre Bundles. 3rd ed.
21 Humphreys. Linear Algebraic Groups.
22 Barnes/Mack. An Algebraic Introduction
to Mathematical Logic.
23 Greub. Linear Algebra. 4th ed.
24 Holmes. Geometric Functional Analysis
and Its Applications.
25 Hewitt/Stromberg. Real and Abstract
Analysis.
26 Manes. Algebraic Theories.
27 Kelley. General Topology.
28 Zariski/Samuel. Commutative Algebra.
Vol.I.
29 Zariski/Samuel. Commutative Algebra.
Vol.II.
30 Jacobson. Lectures in Abstract Algebra I.
Basic Concepts.
3 1 Jacobson. Lectures in Abstract Algebra II.
Linear Algebra.
32 Jacobson. Lectures in Abstract Algebra
III. Theory of Fields and Galois Theory.
33 Hirsch. Differential Topology.
34 Spitzer. Principles of Random Walk.
2nd ed.
35 Alexander/Wermer. Several Complex
Variables and Banach Algebras. 3rd ed.
36 Kelley/Namioka et al. Linear
Topological Spaces.
37 Monk. Mathematical Logic.
38 Grauert/Fritzsche. Several Complex
Variables.
39 Arveson. An Invitation to C* -Algebras.
40 Kemeny/Snell/Knapp. Denumerable
Markov Chains. 2nd ed.
41 Apostol. Modular Functions and Dirichlet
Series in Number Theory.
2nd ed.
42 Serre. Linear Representations of Finite
Groups.
43 Gillman/Jerison. Rings of Continuous
Functions.
44 Kendig. Elementary Algebraic Geometry.
45 Loeve. Probability Theory I. 4th ed.
46 Loeve. Probability Theory II. 4th ed.
47 Moise. Geometric Topology in
Dimensions 2 and 3.
48 Sachs/ W u. General Relativity for
Mathematicians.
49 Gruenberg/Weir. Linear Geometry.
2nd ed.
50 Edwards. Fermat's Last Theorem.
5 1 Klingenberg. a Course in Differential
Geometry.
52 Hartshorne. Algebraic Geometry.
53 Manin. a Course in Mathematical Logic.
54 Graver/Watkins. Combinatorics with
Emphasis on the Theory of Graphs.
55 Brown/Pearcy. Introduction to Operator
Theory I: Elements of Functional
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56 Massey. Algebraic Topology: An
Introduction.
57 Crowell/Fox. Introduction to Knot
Theory.
58 Koblitz. /7-adic Numbers, /?-adic Analysis,
and Zeta-Functions. 2nd ed.
59 Lang. Cyclotomic Fields.
60 Arnold. Mathematical Methods in
Classical Mechanics. 2nd ed.
61 Whitehead. Elements of Homotopy
Theory.
62 Kargapolov/Merlzjakov. Fundamentals
of the Theory of Groups.
63 Bollobas. Graph Theory.
64 Edwards. Fourier Series. Vol. I. 2nd ed.
65 Wells. Differential Analysis on Complex
Manifolds. 2nd ed.
(continued after index)
Bruce E. Sagan
The Symmetric Group
Representations, Combinatorial
Algorithms, and Symmetric Functions
Second Edition
With 31 Figures
Springer
Bruce E. Sagan
Department of Mathematics
Michigan State University
East Lansing, MI 48824-1027
USA
Editorial Board
S. Axler
Mathematics Department
San Francisco State
University
San Francisco, CA 94132
USA
F.W. Gehring
Mathematics Department
East Hall
University of Michigan
Ann Arbor, MI 48109
USA
K.A. Ribet
Mathematics Department
University of California
at Berkeley
Berkeley, CA 94720-3840
USA
Mathematics Subject Classification (2000): 20Cxx, 20C30, 20C32, 05C85
Library of Congress Cataloging-in-Publication Data
Sagan, Bruce Eli.
The symmetric group : representations, combinatorial algorithms, and symmetric
functions / Bruce E. Sagan.
p. cm. — (Graduate texts in mathematics ; 203)
Includes bibliographical references and index.
ISBN 978-1-4419-2869-6 ISBN 978-1-4757-6804-6 (eBook)
DOI 10.1007/978-1-4757-6804-6
1. Representations of groups. 2. Symmetric functions. I. Title. II. Series.
QA171 .S24 2000
512.2— dc21 00-040042
Printed on acid-free paper.
First edition © 1991 by Wadsworth, Inc., Belmont, California.
© 2001 Springer Science+Business Media New York
Originally published by Springer- Verlag New York, Inc. in 2001
Softcover reprint of the hardcover 2nd edition 2001
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Photocomposed copy prepared from the author’s LaTeX files.
987654321
ISBN 978-1-4419-2869-6
SPIN 10770720
Preface to the 2nd Edition
I have been very gratified by the response to the first edition, which has
resulted in it being sold out. This put some pressure on me to come out with
a second edition and now, finally, here it is.
The original text has stayed much the same, the major change being
in the treatment of the hook formula which is now based on the beautiful
Novelli-Pak-Stoyanovskii bijection [NPS 97]. I have also added a chapter on
applications of the material from the first edition. This includes Stanley’s
theory of differential posets [Stn 88, Stn 90] and Fomin’s related concept of
growths [Fom 86, Fom 94, Fom 95], which extends some of the combinatorics
of 5n-representations. Next come a couple of sections showing how groups
acting on posets give rise to interesting representations that can be used to
prove unimodality results [Stn 82]. Finally, we discuss Stanley’s symmetric
function analogue of the chromatic polynomial of a graph [Stn 95, Stn ta].
I would like to thank all the people, too numerous to mention, who pointed
out typos in the first edition. My computer has been severely reprimanded
for making them. Thanks also go to Christian Krattenthaler, Tom Roby,
and Richard Stanley, all of whom read portions of the new material and gave
me their comments. Finally, I would like to give my heartfelt thanks to my
editor at Springer, Ina Lindemann, who has been very supportive and helpful
through various difficult times.
Ann Arbor, Michigan, 2000
V
Preface to the 1st Edition
In recent years there has been a resurgence of interest in representations
of symmetric groups (as well as other Coxeter groups). This topic can be
approached from three directions: by applying results from the general theory
of group representations, by employing combinatorial techniques, or by using
symmetric functions. The fact that this area is the confluence of several
strains of mathematics makes it an exciting one in which to study and work.
By the same token, it is more difficult to master.
The purpose of this monograph is to bring together, for the first time
under one cover, many of the important results in this field. To make the
work accessible to the widest possible audience, a minimal amount of prior
knowledge is assumed. The only prerequisites are a familiarity with elemen-
tary group theory and linear algebra. All other results about representations,
combinatorics, and symmetric functions are developed as they are needed.
Hence this book could be read by a graduate student or even a very bright
undergraduate. For researchers I have also included topics from recent journal
articles and even material that has not yet been published.
Chapter 1 is an introduction to group representations, with special empha-
sis on the methods of use when working with the symmetric groups. Because
of space limitations, many important topics that are not germane to the rest
of the development are not covered. These subjects can be found in any of
the standard texts on representation theory.
In Chapter 2, the results from the previous chapter are applied to the
symmetric group itself, and more highly specialized machinery is developed
to handle this case. I have chosen to take the elegant approach afforded by the
Specht modules rather than working with idempotents in the group algebra.
The third chapter focuses on combinatorics. It starts with the two famous
formulae for the dimensions of the Specht modules: the Frame-Robinson-
Thrall hook formula and the Frobenius- Young determinantal formula. The
centerpiece is the Robinson-Schensted-Knuth algorithm, which allows us to
describe some of the earlier theorems in purely combinatorial terms. A thor-
ough discussion of Schiitzenberger’s jeu de taquin and related matters is in-
cluded.
Chapter 4 recasts much of the previous work in the language of symmet-
ric functions. Schur functions are introduced, first combinatorially as the
generating functions for semistandard tableaux and then in terms of sym-
metric group characters. The chapter concludes with the famous Littlewood-
Richardson and Murnaghan-Nakayama rules.
My debt to several other books will be evident. Much of Chapter 1 is
based on Ledermann’s exquisite text on group characters [Led 77]. Chapter
vii
PREFACE TO THE 1ST EDITION
viii
2 borrows heavily from the monograph of James [Jam 78], whereas Chapter
4 is inspired by Macdonald’s already classic book [Mac 79]. Finally, the third
chapter is a synthesis of material from the research literature.
There are numerous applications of representations of groups, and in par-
ticular of the symmetric group, to other areas. For example, they arise in
physics [Boe 70], probability and statistics [Dia 88], topological graph the-
ory [Whi 84], and the theory of partially ordered sets [Stn 82]. However, to
keep the length of this text reasonable, I have discussed only the connections
with combinatorial algorithms.
This book grew out of a course that I taught while visiting the Universite
du Quebec a Montreal during the fall of 1986. I would like to thank Tequipe
de combinatoire for arranging my stay. I also presented this material in a
class here at Michigan State University in the winter and spring of 1990. I
thank my students in both courses for many helpful suggestions (and those
at UQAM for tolerating my bad French). Francesco Brenti, Kathy Dempsey,
Yoav Dvir, Kathy Jankoviak, and Scott Mathison have all pointed out ways
in which the presentation could be improved. I also wish to express my
appreciation of John Kimmel, Marlene Thom, and Linda Loba at Wadsworth
and Brooks/Cole for their help during the preparation of the manuscript.
Because I typeset this document myself, all errors can be blamed on my
computer.
East Lansing, Michigan, 1991
Contents
Preface to the 2nd Edition v
Preface to the 1st Eition vii
1 Group Representations 1
1.1 Fundamental Concepts 1
1.2 Matrix Representations 4
1.3 G-Modules and the Group Algebra 6
1.4 Reducibility 10
1.5 Complete Reducibility and Maschke’s
Theorem 13
1.6 G-Homomorphisms and Schur’s Lemma 18
1.7 Commutant and Endomorphism Algebras 23
1.8 Group Characters 30
1.9 Inner Products of Characters 33
1.10 Decomposition of the Group Algebra 40
1.11 Tensor Products Again 43
1.12 Restricted and Induced Representations 45
1.13 Exercises 48
2 Representations of the Symmetric Group 53
2.1 Young Subgroups, Tableaux, and Tabloids 53
2.2 Dominance and Lexicographic Ordering 57
2.3 Specht Modules 60
2.4 The Submodule Theorem 63
2.5 Standard Tableaux and a Basis for 5^^ 66
2.6 Garnir Elements 70
2.7 Young’s Natural Representation 74
2.8 The Branching Rule 76
2.9 The Decomposition of 78
2.10 The Semistandard Basis for Hom(S'^, M^) 82
2.11 Kostka Numbers and Young’s Rule 85
2.12 Exercises 86
IX
X
CONTENTS
3 Combinatorial Algorithms 91
3.1 The Robinson-Schensted Algorithm 91
3.2 Column Insertion 95
3.3 Increasing and Decreasing Subsequences 97
3.4 The Knuth Relations 99
3.5 Subsequences Again 102
3.6 Viennot’s Geometric Construction 106
3.7 Schiitzenberger’s Jeu de Taquin 112
3.8 Dual Equivalence 116
3.9 Evacuation 121
3.10 The Hook Formula 124
3.11 The Determinantal Formula 132
3.12 Exercises 133
4 Symmetric Functions 141
4.1 Introduction to Generating Functions 142
4.2 The Hillman-Grassl Algorithm 147
4.3 The Ring of Symmetric Functions 151
4.4 Schur Functions 155
4.5 The Jacobi- Trudi Determinants 158
4.6 Other Definitions of the Schur Function 163
4.7 The Characteristic Map 167
4.8 Knuth’s Algorithm 169
4.9 The Littlewood- Richardson Rule 174
4.10 The Murnaghan-Nakayama Rule 179
4.11 Exercises 185
5 Applications and Generalizations 191
5.1 Young’s Lattice and Differential Posets 191
5.2 Growths and Local Rules 197
5.3 Groups Acting on Posets 204
5.4 Unimodality 208
5.5 Chromatic Symmetric Functions 213
5.6 Exercises 218
Bibliography 223
Index
230
List of Symbols
Symbol
Meaning
Page
A
poset
58
Ak
kth rank of a graded poset
195
As
rank selected poset
204
alternant corresponding to composition fi
164
a\ai
composition a with first part a\ removed
180
a
set partition
215
c
complex numbers
4
ch"
characteristic map
167
Cn
cyclic group of order n
5
Ct
column-stabilizer group of Young tableau t
60
ComX
commutant algebra of representation X
23
C[G]
group algebra of G over C
8
C[S]
vector space generated by set S over C
7
c[W]
ring of formal power series in x over C
142
column insertion operator for integer x
95
cU
Littlewood-Richardson coefficient
175
xig)
character of group element g
30
Xk
character value on conjugacy class K
32
^def
defining character of Sn
31
regular character
31
xIh
restriction of character \ from G to P
45
xtg
induction of character x from hio G
45
D
down operator of a poset
196
length of tt’s longest /c-decreasing subsequence
103
AQ
delta operator applied to tableau Q
121
degX
degree of representation X
4
DesP
the descent set of a tableau P
206
ca
Ath elementary symmetric function
154
nth elementary symmetric function
152
et
polytabloid associated with Young tableau t
61
E{F)
edges of graph T
213
XI
LIST OF SYMBOLS
xii
Symbol Meaning
Page
EndF
endomorphism algebra of module V
23
evQ
the evacuation tableau of Q
122
e
identity element of a group
1
fs
weight generating function of weighted set S
145
number of standard A-tableaux
73
(j)^
character of permutation module
56
G
group
1
9a, B
Garnir element of a pair of sets A, B
70
GLd
dx d complex general linear group
4
GP
generalized permutations
169
GP'
generalized permutations, no repeated columns
172
r
graph
213
hn
nth complete homogeneous symmetric function
152
hx
Ath complete homogeneous symmetric function
154
Hom(P, W)
all module homomorphisms from V to W
22
hij
hooklength of cell {i,j)
124
Hi,j
hook of cell
124
hy
hooklength of cell v
124
Hy
hook of cell v
124
I
identity matrix
23
Id
dx d identity matrix
24
uW
length of tt’s longest /^-increasing subsequence
103
ixiV)
number of stable partitions of V of type A
215
ime
image of 6
21
m
jeu de taquin tableau of P
116
UP)
forward slide on tableau P into cell c
113
UP)
backward slide on tableau P into cell c
113
3q{P)
sequence of forward slides on P into Q
118
UiQ)
sequence of backward slides on Q into P
118
kere
kernel of 6
21
K 9
conjugacy class of a group element g
3
kx
size of Kx
3
Kx
conjugacy class in Sn corresponding to A
3
Kx^
Kostka number
85
Kt
signed column sum for Young tableau t
61
LIST OF SYMBOLS
xiii
Symbol Meaning
Page
/(A)
length (number of parts) of partition A
152
m
leg length of rim hook ^
180
A
integer partition
2
integer composition
67
A'
conjugate or transpose of partition A
103
(Ai, A 2 , . . . , A;)
partition given as a sequence
2
composition given as a sequence
15
Xlfi
skew shape
112
A\e
partition A with rim hook ^ removed
180
m
type of the set partition /?
215
A
algebra of symmetric functions
151
Ai
symmetric functions in 1 variables
163
A"
symmetric functions homogeneous of degree n
152
Mat<j
full d X d matrix algebra
4
m\
Ath monomial symmetric function
151
permutation module corresponding to A
56
P
partial tableau
92
p{n)
number of partitions of n
143
Pn
nth power sum symmetric function
152
Pa
Ath power sum symmetric function
154
PM
P-tableau of permutation tt
93
Pt
chromatic polynomial of graph F
214
P
positive integers
214
7T
permutation
1
7Tp
row word of the Young tableau P
101
reversal of permutation tt
97
7T
top row of generalized permutation tt
169
7T
bottom row of generalized permutation tt
169
n^{P,Q)
Robinson-Schensted map
92
Robinson-Schensted-Knuth map
170
7T ^ — >• (P, g)
dual Robinson-Schensted-Knuth map
172
g(7r)
Q-tableau of permutation tt
93
P(G)
vector space of class functions on G
32
P”
class functions of Sn
167
Rt
row-stabilizer group of Young tableau t
60
Tx
row-insertion operator for the integer x
93
rk
the rank function in a poset
195
XIV
LIST OF SYMBOLS
Symbol Meaning
Page
5a
group of permutations of A
20
sa
Schur function associated with A
155
S\/fi
skew Schur function associated with X/ fi
175
Sx
Young subgroup associated with partition A
54
Specht module associated with partition A
62
sgn(7r)
sign of permutation tt
4
shi
shape of Young tableau t
55
5n
symmetric group on { 1 , 2 ,. ..,n}
1
t
Young tableau
55
T
generalized Young tableau
78
{t]
row tabloid of t
55
M
column tabloid of t
72
T\/i
generalized tableaux, shape A, content /i
78
n.
semistandard tableaux, shape A, content fjL
81
6
homomorphism of modules
18
6t
homomorphism corresponding to tableau T
80
Ot
restriction of 6t to a Specht module
81
U
up operator of a poset
196
vq(P)
vacating tableau for jqiP)
118
v^{Q)
vacating tableau for j^{Q)
118
V{T)
vertices of graph F
213
w-^
orthogonal complement of W
15
wt
weight function
145
X
the set of variables (xi, X 2 , 2 : 3 , . . .}
151
X{9)
matrix of g in the representation X
4
monomial weight of a tableau T
155
x/i
monomial weight of a composition g
155
restriction of representation X from G to H
45
xtg
induction of representation X from H to G
45
Xr
chromatic symmetric function of graph F
214
rim or skew hook
180
Y
Young’s lattice
192
Za
center of algebra A
27
z,
centralizer of group element g
3
Z\
size of Zg where g G Sn has cycle type A
3
Z
integers
204
LIST OF SYMBOLS
XV
Symbol
Meaning
Page
6
minimum of a poset
195
i
maximum of a poset
204
1g
trivial representation of G
5
partition given by multiplicities
2
r\j
equivalence of modules
19
slide equivalence of tableaux
114
♦
r\j
dual equivalence of tableaux
117
K
r\j
Knuth equivalence
100
K*
dual Knuth equivalence
111
P
r\j
P-equivalence
99
Q
r\j
Q-equivalence
111
1
r\j
Knuth relation of the first kind
99
r
r\j
dual Knuth relation of the first kind
111
2
r\j
Knuth relation of the second kind
99
2*
dual Knuth relation of the second kind
111
<
subgroup relation
9
submodule relation
10
lexicographic order on partitions
59
covering relation in a poset
58
>
dominance order on partitions
58
dominance order on tabloids
68
h
is a partition of for integers
53
is a partition of for sets
215
0
direct sum of matrices
13
direct sum of vector spaces
13
0
tensor product of matrices
25
tensor product of representations
43
tensor product of vector spaces
26
inner product of characters \ '0
34
X*
product of characters
168
product of posets
198
GlH
wreath product of groups
212
U
union of tableaux
120
l±)
disjoint union
45
1 • 1
cardinality of a set
3
sum of parts of a partition
54
A
meet operation in a poset
192
V
join operation in a poset
192
Chapter 1
Group Representations
We begin our study of the symmetric group by considering its representations.
First, however, we must present some general results about group represen-
tations that will be useful in our special case. Representation theory can be
couched in terms of matrices or in the language of modules. We consider
both approaches and then turn to the associated theory of characters. All
our work will use the complex numbers as the ground field in order to make
life as easy as possible.
We are presenting the material in this chapter so that this book will be
relatively self-contained, although it all can be found in other standard texts.
In particular, our exposition is modeled on the one in Ledermann [Led 77].
1.1 Fundamental Concepts
In this section we introduce some basic terminology and notation. We pay
particular attention to the symmetric group.
Let G be a group written multiplicatively with identity e. Throughout
this work, G is finite unless stated otherwise. We assume that the reader is
familiar with the elementary properties of groups (cosets, Lagrange’s theorem,
etc.) that can be found in any standard text such as Herstein [Her 64].
Our object of study is the symmetric group, Sn, consisting of all bijections
from {1,2, ... ,n} to itself using composition as the multiplication. The ele-
ments 7T e Sn are called permutations. We multiply permutations from right
to left. (In fact, we compose all functions in this manner.) Thus ttct is the
bijection obtained by first applying cr, followed by tt.
If 7T is a permutation, then there are three different notations we can use
for this element. Two-line notation is the array
_ 1 2 • • • n
^ 7t(1) 7t( 2) ••• 7r(n)
1
2
CHAPTER 1. GROUP REPRESENTATIONS
For example, if tt G 5s is given by
7t(1) = 2, 7t(2) = 3, 7t(3) = 1, 7t(4) = 4, 7t(5) = 5,
then its two-line form is
_ 1 2 3 4 5
’^“ 2 3 1 4 5'
Because the top line is fixed, we can drop it to get one-line notation.
Lastly, we can display tt using cycle notation. Given i G {1,2,. ..,n}, the
elements of the sequence z, 7 t(z), 7T^(z), 7T^(z), . . . cannot all be distinct. Taking
the first power p such that 7 t^(z) = z, we have the cycle
(i,7r(i),7r2(i),...,7rP“H0)-
Equivalently, the cycle (z, j. A:, . . . , /) means that tt sends z to j, j to A:, ... ,
and I back to z. Now pick an element not in the cycle containing z and iterate
this process until all members of {1, 2, . . . ,n} have been used. Our example
from the last paragraph becomes
7t = (1,2,3)(4)(5)
in cycle notation. Note that cyclically permuting the elements within a cycle
or reordering the cycles themselves does not change the permutation. Thus
(1, 2, 3)(4)(5) = (2, 3, 1)(4)(5) = (4)(2, 3, 1)(5) = (4)(5)(3, 1, 2).
A k- cycle., or cycle of length A:, is a cycle containing k elements. The
preceding permutation consists of a 3-cycle and two 1-cycles. The cycle type.,
or simply the type, of tt is an expression of the form
where rrik is the number of cycles of length k in tt. The example permutation
has cycle type
(1^2°,3\4°,5°).
A 1-cycle of tt is called a fixedpoint. The numbers 4 and 5 are fixedpoints in
our example. Fixedpoints are usually dropped from the cycle notation if no
confusion will result. An involution is a permutation such that = e. It is
easy to see that tt is an involution if and only if all of tt’s cycles have length
1 or 2.
Another way to give the cycle type is as a partition. A partition of n is a.
sequence
A = (Ai, A 2 , . . . , A/)
where the A^ are weakly decreasing and Yli=i Ai == Thus k is repeated ruk
times in the partition version of the cycle type of tt. Our example corresponds
to the partition
A = (3,1,1).
1.1. FUNDAMENTAL CONCEPTS
3
In any group G, elements g and h are conjugates if
g = khk~^
for some k e G. The set of all elements conjugate to a given g is called
the conjugacy class of g and is denoted by Kg. Conjugacy is an equivalence
relation, so the distinct conjugacy classes partition G. (This is a set partition,
as opposed to the integer partitions discussed in the previous paragraph.)
Returning to 5^, it is not hard to see that if
7T = (n 1 ^2? • • • 5 ^/) * ’ * (J"rm irn-\-l^ • • • ? ^n)
in cycle notation, then for any a ^ Sn
aira~^ = (x{i2), • • • , cr(ii)) ■ ■ ■ (cr(i„), (r(im+i), • • • , cr(i„)).
It follows that two permutations are in the same conjugacy class if and only
if they have the same cycle type. Thus there is a natural one-to-one corre-
spondence between partitions of n and conjugacy classes of Sn>
We can compute the size of a conjugacy class in the following manner.
Let G be any group and consider the centralizer of ^ G G defined by
Zg = {/i G G : hgh ^ = g},
i.e., the set of all elements that commute with g. Now, there is a bijection
between the cosets of Zg and the elements of Kg, so that
(1-1)
where | • | denotes cardinality. Now let G = Sn and use K\ for Kg when g
has type A.
Proposition 1.1.1 If X = and g ^ Sn has type X, then
\Zg\ depends only on X and
\Zg\ =
Proof. Any h E Zg can either permute the cycles of length i among them-
selves or perform a cyclic rotation on each of the individual cycles (or both) .
Since there are m^! ways to do the former operation and i'^^ ways to do the
latter, we are done. ■
Thus equation (1.1) specializes in the symmetric group to
_ n! _ n\
^ zx l"^imi!2"^2777,2! • • • ’
where k\ = \Nx\-
4
CHAPTER 1. GROUP REPRESENTATIONS
Of particular interest is the conjugacy class of transpositions^ which are
those permutations of the form r = («, j)- The transpositions generate Sn
as a group; in fact, the symmetric group is generated by the adjacent trans-
positions (1,2), (2,3), ..., (n — l,n). If tt = where the ti are
transpositions, then we define the sign of tt to be
sgn( 7 r) = (-I)''.
It can be proved that sgn is well defined, i.e., independent of the particular
decomposition of tt into transpositions. Once this is established, it follows
easily that
sgn(7TC7) = sgn(7r) sgn(cr). (1.3)
As we will see, this is an example of a representation.
1.2 Matrix Representations
A matrix representation can be thought of as a way to model an abstract
group with a concrete group of matrices. After giving the precise definition,
we look at some examples.
Let C denote the complex numbers. Let Mat^^ stand for the set of all
d X d matrices with entries in C. This is called a full complex matrix algebra
of degree d. Recall that an algebra is a vector space with an associative
multiplication of vectors (thus also imposing a ring structure on the space).
The complex general linear group of degree d, denoted by GLd^ is the group
of all X = {xij)dxd ^ Maid that are invertible with respect to multiplication.
Definition 1.2.1 A matrix representation of a group G is a group homomor-
phism
X : G — y GLd‘
Equivalently, to each ^ G G is assigned X{g) G Mat^ such that
1. X{e) = I the identity matrix, and
2. X{gh) - X{g)X{h) for all g,heG,
The parameter d is called the degree^ or dimension^ of the representation and
is denoted by degX. ■
Note that conditions 1 and 2 imply that X{g~^) = X{g)~^^ so these matrices
are in GLd as required.
Obviously the simplest representations are those of degree 1. Our first
two examples are of this type.
Example 1.2.2 All groups have the trivial representation^ which is the one
sending every g G G to the matrix (1). This is clearly a representation because
X{e) = (1) and
X(5)X(/i) = (1)(1) = (1) = X(g/i)
1 .2. MATRIX REPRESENTATIONS
5
for dll g^h e G. We often use 1 g oi* just the number 1 itself to stand for the
trivial representation of G. m
Example 1.2.3 Let us find all one-dimensional representations of the cyclic
group of order n, Cn- Let be a generator of Cn, i.e.,
Cn = •••>5” = e}-
If X{g) = (c), c G C, then the matrix for every element of Cn is determined,
since X(g^) = (c^) by property 2 in the preceding definition. But by property
1 ,
(c")=X( 5 «)^X( 6) = (l),
SO c must be an nth root of unity. Clearly, each such root gives a representa-
tion, so there are exactly n representations of Cn having degree 1.
In particular, let n = 4 and C 4 = {e, The four fourth roots
of unity are l,z, —1, —i. If we let the four corresponding representations be
denoted by X^^\ X^‘^\ X^^\ X^^\ then we can construct a table:
where the entry in row i and column j is X^'^\g^) (matrix brackets omitted).
This array is an example of a character table, a concept we will develop in
Section 1.8. (For representations of dimension 1, the representation and its
character coincide.) Note that the trivial representation forms the first row
of the table.
There are other representations of C4 of larger degree. For example, we
can let
However, this representation is really just a combination of and
In the language of Section 1.5, X is completely reducible with irreducible
components and X^^^. We will see that every representation of Cn can
be constructed in this way using the n representations of degree 1 as building
blocks. ■
Example 1.2.4 We have already met a nontrivial degree 1 representation of
Sn- In fact, equation (1.3) merely says that the map X(7 t) = (sgn(7r)) is a
representation, called the sign representation.
Also of importance is the defining representation of which is of degree
n. If 7T G <Sn, then we let X(7 t) = {Xij)nxn^ where
-r . . = / ^
\ 0 otherwise.
6
CHAPTER!. GROUP REPRESENTATIONS
The matrix X(7 t) is called a permutation matrix^ since it contains only zeros
and ones, with a unique one in every row and column. The reader should
verify that this is a representation.
In particular, consider S 3 with its permutations written in cycle notation.
Then the matrices of the defining representation are
/100\ /010\
X(e) =010, X( (1,2) ) = 1 0 0 ,
\001/ \001/
/001\ /100\
X{ (1,3) ) = 0 1 0 , X( (2,3) ) = 0 0 1 ,
\100/ \010/
/001\ /010\
X( (1,2,3))= 10 0, X( (1,3,2))= 001.-
\010/ \100/
1.3 G-Modules and the Group Algebra
Because matrices correspond to linear transformations, we can think of rep-
resentations in these terms. This is the idea of a G-module.
Let F be a vector space. Unless stated otherwise, all vector spaces will
be over the complex numbers and of finite dimension. Let GL{V) stand for
the set of all invertible linear transformations of V to itself, called the general
linear group of V. If dim V = d, then GL{V) and GLd are isomorphic as
groups.
Definition 1.3.1 Let U be a vector space and G be a group. Then U is a
G-module if there is a group homomorphism
p:G-^GL{V).
Equivalently, U is a G-module if there is a multiplication, gv, of elements of
V by elements of G such that
1. gv € U,
2. g{cv + dw) = c{gw) -h d(^w),
3. {gh)w = ^(hv), and
4. ev = V
for all h G G; V, w e U; and scalars c, d G C. ■
1.3. G-MODULES AND THE GROUP ALGEBRA
7
In the future, “G-module” will often be shortened to “module” when no
confusion can result about the group involved. Other words involving G- as
a prefix will be treated similarly. Alternatively, we can say that the vector
space V carries a representation of G.
Let us verify that the two parts of the definition are equivalent. In fact,
we are just using gv as a shorthand for the application of the transformation
p{g) to the vector v. Item 1 says that the transformation takes V to itself; 2
shows that the map is linear; 3 is property 2 of the matrix definition; and 4 in
combination with 3 says that g and g~^ are inverse maps, so all transforma-
tions are invertible. Although it is more abstract than our original definition
of representation, the G-module concept lends itself to cleaner proofs.
In fact, we can go back and forth between these two notions of represen-
tation quite easily. Given a matrix representation X of degree d, let V be
the vector space of all column vectors of length d. Then we can multiply
V eV by g E G using the definition
gv X{g)v,
where the operation on the right is matrix multiplication. Conversely, if V is
a G-module, then take any basis B for V. Thus X{g) will just be the matrix
of the linear transformation g e G in the basis B computed in the usual way.
We use this correspondence extensively in the rest of this book.
Group actions are important in their own right. Note that if S is any
set with a multiplication by elements of G satisfying 1, 3, and 4, then we
say G acts on S. In fact, it is always possible to take a set on which G
acts and turn it into a G-module as follows. Let S = {si, S 2 ? • • • , 5^} and let
CS == C{si,S 2 , . . . ,s^} denote the vector space generated by S over C; i.e.,
CS consists of all the formal linear combinations
Cl Si C 2 S 2 + • • • + CnSji,
where c^ G C for all i. (We put the elements of S in boldface print when they
are being considered as vectors.) Vector addition and scalar multiplication in
CS are defined by
(CiSi + C2S2 + • • • + Cn^n) + (diSi + ^282 + * • * + d^Sn)
= (ci -f di)si -h (C 2 + d 2 )S 2 + • • • + {Cji -h dn)Sn
and
C (ciSi -h C2S2 H h CnSn) = (cCi)Si -f (CC2)S2 H h (cCn)Sn,
respectively. Now the action of G on 5 can be extended to an action on CS
by linearity:
g (ciSi -f C2S2 + • • • + CnSn) = Ci( 5 rSi) -h 02(^82) + * * * + Cn{gSn)
for all g ^ G. This makes CS into a G-module of dimension \S\.
8
CHAPTER 1. GROUP REPRESENTATIONS
Definition 1.3.2 If a group G acts on a set 5, then the associated module
CS is called the permutation representation associated with S. Also, the
elements of S form a basis for CS called the standard basis, m
All the following examples of G-modules are of this form.
Example 1.3.3 Consider the symmetric group Sn with its usual action on
5 = {1,2, . . . ,n}. Now
CS = {cil -h C 22 H h Cnii : Ci eC for all i}
with the action
7t(ci 1 + C22 H h C^n) = ci7t(1) + C27t( 2) H h Cn 7r(n)
for all 7T G Sn-
To make things more concrete, we can select a basis and determine the
matrices X{tt) for tt G 5^ in that basis. Let us consider 53 and use the
standard basis (1, 2, 3}. To find the matrix for tt = (1, 2), we compute
(1,2)1 = 2; (1,2)2 -1; (1,2)3 = 3;
and so
/ 0 1 0
A( (1,2) )= 10 0
\ 0 0 1
If the reader determines the rest of the matrices for 5a, it will be noted that
they are exactly the same as those of the defining representation. Exam-
ple 1.2.4. It is not hard to show that the same is true for any n; i.e., this is
merely the module approach to the defining representation of 5n- ■
Example 1.3.4 We now describe one of the most important representations
for any group, the (left) regular representation. Let G be an arbitrary group.
Then G acts on itself by left multiplication: if ^ G G and h E S = G, then the
action of g on /i, gh^ is defined as the usual product in the group. Properties 1,
3, and 4 now follow, respectively, from the closure, associativity, and identity
axioms for the group.
Thus if G = {^ 1 , ^ 2 , • • • , ^n}, then we have the corresponding G-module
C[G] = {cigi + C 2 g 2 H 1- C„g„ : Ci G C for all i},
which is called the group algebra of G. Note the use of square brackets to
indicate that this is an algebra, not just a vector space. The multiplication is
gotten by letting g^gj = gk in C[G] if gigj = gk in G, and linear extension.
Now the action of G on the group algebra can be expressed as
9 (cigi + C 2 g 2 H + c„g„) = Ci(ggi) + C2(gg2) H + C„(gg„)
for all g G G. The group algebra will furnish us with much important combi-
natorial information about group representations.
1.3. G-MODULES AND THE GROUP ALGEBRA
9
Let us see what the regular representation of the cyclic group G4 looks
like. First of all,
C[C4] = {ci€ + C2g + C3g^ H- C4g^ : Q G C for all i].
We can easily find the matrix of in the standard basis:
= p^g = g^, g^g^ = e, ^V=g-
Thus
X{g^)
/ 0 0 1 0 \
0 0 0 1
10 0 0
\ 0 1 0 0 y
Computing the rest of the matrices would be a good exercise. Note that
they are all permutation matrices and all distinct. In general, the regular
representation of G gives an embedding of G into the symmetric group on
|G| elements. The reader has probably seen this presented in a group theory
course, in a slightly different guise, as Cayley’s theorem [Her 64, pages 60-61].
Note that if G acts on any V, then so does C[G|. Specifically, if cigi H-
C2g2 + • • • H- Cngn ^ C[G] and V G F, then we can define the action
(cigl + C 2 g 2 + h Cngn)v = Ci(^iv) + 02(^2 v) + • * • + Cn(^nV).
In fact, we can extend the concept of representation to algebras: A represen-
tation of an algebra A is an algebra homomorphism from A into GL{V). In
this way, every representation of a group G gives rise to a representation of its
group algebra C[G]. For a further discussion of representations of algebras,
see the text of Curtis and Reiner [C-R 66]. ■
Example 1.3.5 Let group G have subgroup iJ, written H < G. A gen-
eralization of the regular representation is the (left) coset representation
of G with respect to H. Let gi, g2^ • • • 1 9k be a transversal for if; i.e.,
R = {^1 if, ^2^5 • • • ^9kH} is a complete set of disjoint left cosets for if in
G. Then G acts on R by letting
g{giH) = {ggi)H
for all g E G. The corresponding module
CR = {cigiH H- C2g2H -I h CfcgfcH : Ci eC for all i}
inherits the action
g (cigiH CfcgfcH) = ci(ggiH) -h • • • + Cfc(ggfcH).
Note that if if = G, then this reduces to the trivial representation. At
the other extreme, when if = {e}, then R — G and we obtain the regular
10
CHAPTER!. GROUP REPRESENTATIONS
representation again. In general, representation by cosets is an example of an
induced representation, an idea studied further in Section 1.12.
Let us consider G = S 3 and H = {e, (2, 3)}. We can take
n = {H, (1,2)//, (1,3)//}
and
CW = {ciH + C 2 ( 1 , 2 )H + C3(1,3)H : G C for all z}.
Computing the matrix of (1,2) in the standard basis, we obtain
(1,2)H = (1,2)H, (1,2)(1,2)H = H, (1, 2)(1, 3)H = (1, 3, 2)H = (1, 3)H,
SO that
X((l,2)) =
0 1 0
1 0 0
0 0 1
After finding a few more matrices, the reader will become convinced that we
have rediscovered the defining representation yet again. The reason for this
is explained when we consider isomorphism of modules in Section 1.6. ■
1.4 Reducibility
An idea pervading all of science is that large structures can be understood by
breaking them up into their smallest pieces. The same thing is true of repre-
sentation theory. Some representations are built out of smaller ones (such as
the one at the end of Example 1.2.3), whereas others are indivisible (as are
all degree one representations). This is the distinction between reducible and
irreducible representations, which we study in this section. First, however,
we must determine precisely what a piece or subobject means in this setting.
Definition 1.4.1 Let F be a G- module. A submodule of V is a, subspace W
that is closed under the action of G, i.e.,
w eW => gw G W for all g £ G.
We also say that W is a G-invariant subspace. Equivalently, W is a subset of
V that is a G-module in its own right. We write W < V iiW is a submodule
of V. m
As usual, we illustrate the definition with some examples.
Example 1.4.2 Any G-module, F, has the submodules W = V as well as
W = {0}, where 0 is the zero vector. These two submodules are called trivial.
All other submodules are called nontrivial, m
1.4. REDUCIBILITY
11
Example 1.4.3 For a nontrivial example of a submodule, consider G — Sm
n > 2, and V = C{l,2,...,n} (the defining representation). Now take
VF = C{l4-2 + --- + n} - {c(l + 2 + --- + n) : c G C};
i.e., W is the one-dimensional subspace spanned by the vector 1-f 2H hn.
To check that W is closed under the action of 5n, it suffices to show that
7TW G W for all w in some basis for W and all tt G <Sn.
(Why?) Thus we need to verify only that
7t(1 + 2 + • • • + n) G W
for each tt G «Sn. But
7t(1 + 2 H h n) == 7t(1) -f 7t(2) H h 7r(n)
= 1 + 2H + nGW,
because applying tt to {1, 2, . . . ,n} just gives back the same set of numbers
in a different order. Thus IT is a submodule of V that is nontrivial since
dim W = 1 and dim V = n > 2.
Since VF is a module for G sitting inside F, we can ask what representation
we get if we restrict the action of G to IT. But we have just shown that every
7T G 5n sends the basis vector 1 + 2 + • • • + n to itself. Thus X(7t) = (1) is the
corresponding matrix, and we have found a copy of the trivial representation
in C{1,2, . . . ,n}. In general, for a vector space IT of any dimension, if G
fixes every element of IT, we say that G acts trivially on W. m
Example 1.4.4 Next, let us look again at the regular representation. Sup-
pose G = {gi,g 2 ^ • • • ’,9n} with group algebra V = C[G]. Using the same idea
as in the previous example, let
w = c[gi +g 2 H +gn],
the one-dimensional subspace spanned by the vector that is the sum of all
the elements of G. To verify that IT is a submodule, take any g E G and
compute:
aisi + g 2 H h g„) = 5gl + 5g2 H \-9En
= gl + g2 H + gn^W,
because multiplying by g merely permutes the elements of G, leaving the sum
unchanged. As before, G acts trivially on FF.
The reader should verify that if G = <Sn, then the sign representation can
also be recovered by using the submodule
w" = C[E^es„ sgn(7T) tt ] ■
12
CHAPTER 1. GROUP REPRESENTATIONS
We now introduce the irreducible representations that will be the building
blocks of all the others.
Definition 1.4.5 A nonzero G-module V is reducible if it contains a non-
trivial submodule W . Otherwise, V is said to be irreducible. Equivalently, V
is reducible if it has a basis B in which every ^ G G is assigned a block matrix
of the form
A{g)
B{g) '
0
C(g) ,
(1.4)
where the A{g) are square matrices, all of the same size, and 0 is a nonempty
matrix of zeros. ■
To see the equivalence, suppose V of dimension d has a submodule W of
dimension /, 0 < / < d. Then let
e = {wi, W2, . . . , W/, V/+1, V/+2, . . . , Vd},
where the first / vectors are a basis for W. Now we can compute the matrix
of any g E G with respect to the basis B. Since W is a submodule, gwi G W
for all z, 1 < i < /. Thus the last d — f coordinates of gWi will all be zero.
That accounts for the zero matrix in the lower left corner of X{g). Note that
we have also shown that the A{g), g E are the matrices of the restriction
of G to W . Hence they must all be square and of the same size.
Conversely, suppose each X{g) has the given form with every A{g) being
f X f. Let V = C^ and consider
W = C{ei,e2,...,e/},
where is the column vector with a 1 in the zth row and zeros elsewhere
(the standard basis for CA). Then the placement of the zeros in X{g) assures
us that X{g)ei G W for 1 < z < / and all g E G. Thus W is a G-module,
and it is nontrivial because the matrix of zeros is nonempty.
Clearly, any epresentation of degree 1 is irreducible. It seems hard to de-
termine when a representation of greater degree will be irreducible. Certainly,
checking all possible subspaces to find out which ones are submodules is out
of the question. This unsatisfactory state of affairs will be remedied after we
discuss inner products of group characters in Section 1.9.
From the preceding examples, both the defining representation for Sn and
the group algebra for an arbitrary G are reducible if rz > 2 and |G| > 2, respec-
tively. After all, we produced nontrivial submodules. Let us now illustrate
the alternative approach via matrices using the defining representation of S 3 .
We must extend the basis {1 4- 2 -f 3} for W to a basis for V = C{1, 2, 3}.
Let us pick
6 = {1 + 2 + 3, 2, 3}.
Of course, X{e) remains the 3x3 identity matrix. To compute X{ (1,2) ),
we look at (1, 2)’s action on our basis:
(l,2)(l + 2 + 3) = 1 + 2 + 3, (l,2)2 = l = (l + 2 + 3)-2-3, (1, 2)3 = 3.
1.5. COMPLETE REDUCIBILITY AND MASCHKE’S THEOREM 13
So
/I 10
X((l,2))= 0-10
\ 0 -1 1
The reader can do the similar computations for the remaining four elements
of S 3 to verify that
1 0 1
X{ (1,3) ) = I 0 1 -1
0 0-1
1
-1
X{ (2,3))
X( (1,2,3))
1 0
0 0
X{ (1,3,2) ) =
0 1 -1
Note that all these matrices have the form
' 1
*
*
0
*
*
^ 0
*
*
The one in the upper left corner comes from the fact that S 3 acts trivially on
W.
1.5
Complete Reducibility and Maschke’s
Theorem
' A{9)
0 '
0
B{g) ,
It would be even better if we could bring the matrices of a reducible G-module
to the block diagonal form
X{g)
for all g E G. This is the notion of a direct sum.
Definition 1.5.1 Let F be a vector space with subspaces U and W. Then
V is the (internal) direct sum of U and written V = U ii every
V GV can be written uniquely as a sum
V = u + w, u e U, w G W.
If V is a G-module and C/, W are G-submodules, then we say that U and W
are complements of each other.
If X is a matrix, then X is the direct sum of matrices A and written
X = A® if X has the block diagonal form
' A
0 '
0
B
14
CHAPTER 1. GROUP REPRESENTATIONS
To see the relationship between the module and matrix definitions, let V
be a G-module with V = U ^ W, where U,W <V. Since this is a direct sum
of vector spaces, we can choose a basis for V
B = {Ui,U2,...,Uf, W/+1 , W/+2, . . . , Wd}
such that {ui, U2, . . . , u/} is a basis for U and {w/+i, wy+2, . . . , w^} is a
basis for W. Since U and W are submodules, we have
gui G U and gWj G W
for all G G, G C/, wj eW. Thus the matrix of any p G G in the basis B
is
A(g)
0
0
Big) ,
where A{g) and B{g) are the matrices of the action of G restricted to U and
W, respectively.
Returning to the defining representation of ^3, we see that
V - C{1, 2, 3} = C{1 + 2 + 3} 0 C{2, 3}
as vector spaces. But while C{1 0 2 + 3} is an tSa-submodule, C{2,3} is
not (e.g., (1,2)2 = 1 ^ C{2,3}). So we need to find a complement for
C{1 + 2 + 3}, i.e., a submodule U such that
C{1,2,3} =C{l + 2 + 3}0C/.
To find a complement, we introduce an inner product on C{1, 2, 3}. Given
any two vectors i,j in the basis {1,2,3}, let their inner product be
(i,j) = Sij, (1.5)
where Sij is the Kronecker delta. Now we extend by linearity in the first
variable and conjugate linearity in the second to obtain an inner product on
the whole vector space. Equivalently, we could have started out by defining
the product of any two given vectors v = al + 62 + c3, w = xl + y2 + z3 as
(v, w) = ax by cz,
with the bar denoting complex conjugation. The reader can check that this
definition does indeed satisfy all the axioms for an inner product. It also
enjoys the property that it is invariant under the action of G:
{gv,gw) = (v, w) for all ^ G G and v, w G F. (1.6)
To check invariance on V, it suffices to verify (1.6) for elements of a basis.
But if 7T G <Ss, then
(ttI, TTj) Sij (bj)
1.5. COMPLETE REDUCIBILITY AND MASCHKE’S THEOREM 15
where the middle equality holds because tt is a bijection.
Now, given any inner product on a vector space V and a subspace W, we
can form the orthogonal complement:
= {v G F : (v, w) = 0 for all w eW}.
It is always true that V = W ® W-^. When W <V and the inner product is
G-invariant, we can say more.
Proposition 1.5.2 Let V be a G -module, W a submodule, and (*, •) an inner
product invariant under the action of G. Then is also a G-submodule.
Proof. We must show that for all g e G and u G W-^ we have G W-^.
Take any w G W; then
(^u,w) == {g~^gu,g~^w) (since (•,•) is invariant)
= {u,g~^w) (properties of group action)
— 0. (u G W-^, and g~^w G W
since W is a submodule)
Thus W-^ is closed under the action of G. m
Applying this to our running example, we see that
C{l + 2 + 3}-^ - {v-al + 62 + c3 : (v,l + 2 + 3) =0}
= -[v = ul 52 -f" c3 : a b c = 0}.
To compute the matrices of the direct sum, we choose the bases {1 + 2 + 3}
for C(1 + 2 + 3}, and (2-1, 3-1} for C(1 + 2 + 3}-^. This produces the
matrices
X{e) =
1 0 0
0 1 0
0 0 1
X( (1,2)) =
1 0 0
0 -1 -1
0 0 1
X( (1,3) )
1 0 0
0 1 0
0 -1 -1
X( (2,3) ) =
1 0 0
0 0 1
0 1 0
(1,2,3))
JC( (1,3,2)) =
These are indeed all direct matrix sums of the form
X{g)
' A{g)
0 0 '
0
0
B{g)
Of course, A{g) is irreducible (being of degree 1), and we will see in Section 1.9
that B{g) is also. Thus we have decomposed the defining representation of
53 into its irreducible parts. The content of Maschke’s theorem is that this
can be done for any finite group.
16
CHAPTER 1. GROUP REPRESENTATIONS
Theorem 1.5.3 (Maschke’s Theorem) Let G be a finite group and let V
be a nonzero G-module. Then
where each W^'‘^ is an irreducible G-submodule ofV.
Proof. We will induct on d = dimK. If d = 1, then V itself is irreducible
and we are done (A; = 1 and = V). Now suppose that d > 1. If F is
irreducible, then we are finished as before. If not, then V has a nontrivial
G-submodule, W. We will try to construct a submodule complement for W
as we did in the preceding example.
Pick any basis B ~ {vi, V 2 , . . . , v^} for V. Consider the unique inner
product that satisfies
(vi,Vj) =
for elements of B. This product may not be G-invariant, but we can come up
with another one that is. For any v, w G we let
(v,w)' = '^{gv,gw).
g€G
We leave it to the reader to verify that (•, •)' satisfies the definition of an
inner product. To show that it is G-invariant, we wish to prove
(/iv,/iw)' = (v, w)'
for all h e G and v, w G 1^. But
EseG(5/iv,5/iw)
= E/€g(/v,/w)
= (v,w)'
(definition of (•, •)')
(as g varies over G, so does / = gh)
(definition of (•, •)')
as desired.
If we let
= {v eV : (v, w)' = 0},
then by Proposition 1.5.2 we have that W-^ is a G-submodule of V with
Y = W^W^.
Now we can apply induction to W and W-^ to write each as a direct sum of
irreducibles. Putting these two decompositions together, we see that V has
the desired form. ■
As a corollary, we have the matrix version of Maschke’s theorem. Here
and in the future, we often drop the horizontal and vertical lines indicating
block matrices. Our convention of using lowercase letters for elements and
uppercase ones for matrices should avoid any confusion.
1.5. COMPLETE REDUCIBILITY AND MASCHKE’S THEOREM 17
Corollary 1.5.4 Let G be a finite group and let X he a matrix representation
of G of dimension d > 0. Then there is a fixed matrix T such that every
matrix X{g), g G G^ has the form
\
/
Proof. Let V = with the action
gv = X{g)v
for all ^ G G and v eV. By Maschke’s theorem,
each being irreducible of dimension, say, di. Take a basis B for V such
that the first d\ vectors are a basis for the next ^2 are a basis for
etc. The matrix T that transforms the standard basis for into B now does
the trick, since conjugating by T just expresses each X{g) in the new basis
B. m
Representations that decompose so nicely have a name.
Definition 1.5.5 A representation is completely reducible if it can be written
as a direct sum of irreducibles. ■
So Maschke’s theorem could be restated:
Every representation of a finite group having positive dimension is
completely reducible.
We are working under the nicest possible assumptions, namely, that all
our groups are finite and all our vector spaces are over C. We will, however,
occasionally attempt to indicate more general results. Maschke’s theorem
remains true if C is replaced by any field whose characteristic is either zero or
prime to |G|. For a proof in this setting, the reader can consult Ledermann
[Led 77, pages 21-23].
However, we can not drop the finiteness assumption on G, as the following
example shows. Let R"*" be the positive real numbers, which are a group under
multiplication. It is not hard to see that letting
for all r G defines a representation. The subspace
TX{g)T-^
( 0
0 X(2)(g)
Vo 0 ••• XW(g)
where each X^^^ is an irreducible matrix representation of G.
18
CHAPTER 1. GROUP REPRESENTATIONS
is invariant under the action of G. Thus if X is completely reducible, then
must decompose as the direct sum of W and another one-dimensional
submodule. By the matrix version of Maschke’s theorem, there exists a fixed
matrix T such that
rxMT- = (“=<;>
for all r G R"^. Thus x(r) and y{r) must be the eigenvalues of X(r), which
are both 1. But then
X{r)=T-^( I ^^)t =
1 0
0 1
for all r G R“*“, which is absurd.
1.6 G-Homomorphisms and Schur’s Lemma
We can learn more about objects in mathematics (e.g., vector spaces, groups,
topological spaces) by studying functions that preserve their structure (e.g.,
linear transformations, homomorphisms, continuous maps). For a G-module,
the corresponding function is called a G-homomorphism.
Definition 1.6.1 Let V and W be G-modules. Then a G-homomorphism
(or simply a homomorphism) is a linear transformation 6 : V W such that
6 *( 9 v ) = gOiv)
for all ^ G G and v G F. We also say that 9 preserves or respects the action
of G. ■
We can translate this into the language of matrices by taking bases B and C
for V and W, respectively. Let X{g) and Y{g) be the corresponding matrix
representations. Also, take T to be the matrix of 6 in the two bases B and C.
Then the G-homomorphism property becomes
TX(g)v = Y(g)Tv
for every column vector v and g ^ G. But since this holds for all v, we must
have
TX{g) = Y{g)T for all geG, (1.7)
Thus having a G-homomorphism 6 is equivalent to the existence of a matrix T
such that (1.7) holds. We will often write this condition simply as TA — YT.
As an example, let G == 5^, = C{v} with the trivial action of and let
W = C{1, 2 , . , . , n} with the defining action of Sn- Define a transformation
9:V Why
6{v) = 1 + 2 + . • • + n
1.6. G-HOMOMORPHISMS AND SCHUR’S LEMMA
19
and linear extension; i.e.,
9{cv) = c(l + 2 H 1- n)
for all c G C. To check that 0 is a G- homomorphism, it suffices to check that
the action of G is preserved on a basis of V. (Why?) But for all tt € cS„,
n n
6{'Kw) — 6{w) = ^ i = 7T ^ i = 7T0(v).
2=1
In a similar vein, let G be an arbitrary group acting trivially onV = C{v},
and let W — C[G] be the group algebra. Now we have the G-homomorphism
9 :V -^W given by extending
= XI e
geG
linearly.
If G = we can also let G act onV = G{v} by using the sign represen-
tation:
7TU = sgn(7r)u
for all 7T G <Sn and u eV. Keeping the usual action on the group algebra, the
reader can verify that
= XZ sgn(7r)7T
7rG<Sn
extends to a G-homomorphism from V to W.
It is clearly important to know when two representations of a group are
different and when they are not (even though there may be some cosmetic
differences). For example, two matrix representations that differ only by a
basis change are really the same. The concept of G-equi valence captures this
idea.
Definition 1.6.2 Let V and W be modules for a group G. A G -isomorphism
is a G-homomorphism 6 : V W that is bijective. In this case we say that
V and W are G-isomorphic^ or G-equivalent, written V = W. Otherwise we
say that V and W are G -inequivalent ■
As usual, we drop the G when the group is implied by context.
In matrix terms, 6 being a bijection translates into the corresponding
matrix T being invertible. Thus from equation (1.7) we see that matrix
representations X and T of a group G are equivalent if and only if there
exists a fixed matrix T such that
Y{g) = TX{g)T-^
for all ^ G G. This is the change-of-basis criterion that we were talking about
earlier.
20
CHAPTER 1. GROUP REPRESENTATIONS
Example 1.6.3 We are now in a position to explain why the coset rep-
resentation of S3 at the end of Example 1.3.5 is the same as the defining
representation. Recall that we had taken the subgroup H = {e, (2,3)} C S3
giving rise to the coset representation module CH, where
{l,2)H,
Given any set let Sa be the symmetric group on A, i.e., the set of all
permutations of T. Now the subgroup H can be expressed as an (internal)
direct product
= {(1)(2)(3), (l)(2,3)} = {(l)}x{(2)(3), (2,3)} = <S{i} x<S{ 2 , 3 }- (1-8)
A convenient device for displaying such product subgroups of Sn is the
tabloid. Let A = (Ai, A 2 , . . . , Af) be a partition, as discussed in Section 1.1. A
Young tabloid of shape A is an array with I rows such that row i contains A^
integers and the order of entries in a row does not matter. To show that each
row can be shuffled arbitrarily, we put horizontal lines between the rows. For
example, if A = (4, 2, 1), then some of the possible Young tabloids are
3 1 4 1 3 1 1 4 9 5 3 4
5 9 = 9 5 ^21
2 2 1
Equation (1.8) says that H consists of all permutations in ^3 that permute
the elements of the set { 1 } among themselves (giving only the permutation
( 1 )) and permute the elements of {2,3} among themselves (giving (2) (3) and
(2,3)). This is modeled by the tabloid
2 3
1
since the order of 2 and 3 is immaterial but 1 must remain fixed. The complete
set of tabloids of shape A = (2, 1) whose entries are exactly 1, 2, 3 is
c_ f 2 3 13 1 2 \
\j_ ’ _2_ ’ _3_ J '
Furthermore, there is an action of any tt G ^3 on 5 given by
_ i 3 ^ 7r(i) 7r(i)
k 7r(k)
Thus it makes sense to consider the map 6 that sends
H A 4—^
A (1,3)^-^==-!^
(1,3)H
1
3
1.6. G-HOMOMORPHISMS AND SCHUR’S LEMMA
21
By linear extension, 6 becomes a vector space isomorphism from CH to CS.
In fact, we claim it is also a G-isomorphism. To verify this, we can check that
the action of each tt E S 3 is preserved on each basis vector in H. For example,
if 7T = (1, 2) and H G then
0((1,2)H) = ^((1,2)H) = = (1,2)^-^ = (1,2)0(H).
Thus
cn ^ CS. (1.9)
Another fact about the tabloids in our set S is that they are completely
determined by the element placed in the second row. So we have a natural
map, Tj, between the basis {1,2,3} for the defining representation and S,
namely.
1
r; 2 3
T—
2
4
1 3
2
3
4
1 2
3
Now T] extends by linearity to a G-isomorphism from C{1, 2, 3} to CS. This,
in combination with equation (1.9), shows that CW and C{1, 2, 3} are indeed
equivalent.
The reader may feel that we have taken a long and winding route to get
to the final iSs-isomorphism. However, the use of Young tabloids extends far
beyond this example. In fact, we use them to construct all the irreducible
representations of Sn in Chapter 2. ■
We now return to the general exposition. Two sets usually associated
with any map of vector spaces 6 \ V -^W are its kernel.,
kerO = {v eV : 6 {v) = 0 },
where 0 is the zero vector, and its image,
im 6 = {w eW : w = 6 {v) for some v G V}.
When ^ is a G-homomorphism, the kernel and image have nice structure.
Proposition 1.6.4 Let 6 :V W be a G-homomorphism. Then
1 . ker0 is a G -submodule ofV, and
2. im 9 is a G-submodule ofW.
22
CHAPTER 1. GROUP REPRESENTATIONS
Proof. We prove only the first assertion, leaving the second one for the reader.
It is known from the theory of vector spaces that ker 0 is a subspace of V since
6 is linear. So we only need to show closure under the action of G. But if
V G ker 6, then for any g £ G,
^( 5 V) = gO{v)
{6 is a G-homomorphism)
= gO
(v 6 ker 9)
= 0,
and so gv £ ker 6, as desired. ■
It is now an easy matter to prove Schur’s lemma, which characterizes G-
homomorphisms of irreducible modules. This result plays a crucial role when
we discuss the commutant algebra in the next section.
Theorem 1.6.5 (Schur’s Lemma) Let V and W be two irreducible G-
modules. If 0 : V W is a G -homomorphism, then either
1. 6 is a G -isomorphism, or
2. 6 is the zero map.
Proof. Since V is irreducible and ker^ is a submodule (by the previous
proposition), we must have either ker^ = {0} or ker^ = V. Similarly, the
irreducibility of W implies that im0 = {0} or W. If ker^ = V or im0 = {0},
then 6 must be the zero map. On the other hand, if ker 6 = {0} and im9 = W,
then we have an isomorphism. ■
It is interesting to note that Schur’s lemma continues to be valid over
arbitrary fields and for infinite groups. In fact, the proof we just gave still
works. The matrix version is also true in this more general setting.
Corollary 1.6.6 Let X and Y be two irreducible matrix representations of
G. IfT is any matrix such that TX{g) = Y{g)T for all g £ G, then either
1. T is invertible, or
2. T is the zero matrix. ■
We also have an analogue of Schur’s lemma in the case where the range
module is not irreducible. This result is conveniently expressed in terms of
the vector space Hom(V, W) of all G-homomorphisms from V to W.
Corollary 1.6.7 Let V and W be two G-modules with V being irreducible.
Then dim Hom(V, W) = 0 if and only ifW contains no submodule isomorphic
to V. m
When the field is C, however, we can say more. Suppose that T is a matrix
such that
TX{g) = X{g)T (1.10)
1 . 7. COMMUTANT AND ENDOMORPHISM ALGEBRAS
23
for all g £ G. It follows that
{T-cI)X = X{T-cI),
where I is the appropriate identity matrix and c G C is any scalar. Now C is
algebraically closed, so we can take c to be an eigenvalue of T. Thus T — cl
satisfies the hypothesis of Corollary 1.6.6 (with X = Y) and is not invertible
by the choice of c. Our only alternative is that T — cl = 0. We have proved
the following result:
Corollary 1.6.8 Let X be an irreducible matrix representation ofG over the
complex numbers. Then the only matrices T that commute with X{g) for all
g E G are those of the form T — cl — i.e., scalar multiples of the identity
matrix, m
1.7 Commutant and Endomorphism Algebras
Corollary 1.6.8 suggests that the set of matrices that commute with those of
a given representation are important. This corresponds in the module setting
to the set of G-homomorphisms from a G-module to itself. We characterize
these sets in this section. Extending these ideas to homomorphisms between
different G-modules leads to a useful generalization of Corollary 1.6.7 (see
Corollary 1.7.10).
Definition 1.7.1 Given a matrix representation X \ G GLd, the corre-
sponding commutant algebra is
ComX = {T e Matd : TX{g) = X{g)T for all g G G},
where Mat^ is the set of all dx d matrices with entries in C. Given a G-module
V, the corresponding endomorphism algebra is
Endy — {0 :V : 0isa G-homomorphism}. ■
It is easy to check that both the commutant and endomorphism algebras do
satisfy the axioms for an algebra. The reader can also verify that if F is a
G-module and X is a corresponding matrix representation, then End V and
Com X are isomorphic as algebras. Merely take the basis B that produced X
and use the map that sends each 6 G End V to the matrix T of 0 in the basis
B. Let us compute ComX for various representations X.
Example 1.7.2 Suppose that X is a matrix representation such that
where X^^^X^^^ are inequivalent and irreducible of degrees di,d 2 , respec-
tively. What does Com X look like?
24
CHAPTER 1. GROUP REPRESENTATIONS
Suppose that
T =z ( ^1,2 ^
~ V ^2,1 T2,2 J
is a matrix partitioned in the same way as X. U T X = XT , then we can
multiply out each side to obtain
V T2’iX(1) T2’2X(2) ; “ V X(2)T2’i X<2)T2’2 J ‘
Equating corresponding blocks we get
Ti,2X(2)
T2,2X(2)
= X<2)T2,2.
Using Corollaries 1.6.6 and 1.6.8 along with the fact that X^^^ and X^“^^ are
inequivalent, these equations can be solved to yield
^1,1 = Cl 7^1 , Ti, 2 = ^2,1 = 0, T2,2 = C2/d2?
where ci,C 2 G C and Idi^Id 2 identity matrices of degrees di,d 2 - Thus
T = f ^ \
V 0 ^2/d. ; ■
We have shown that when X = 0 with X^^^ ^ and irre-
ducible, then
ComX == {cild^ 0C27d2 : ci,C2 G C},
where di = degX^^\ c ?2 = degX^^^.B
In general, if X = ®i=iX^'^\ where the X^'^^ are pairwise inequivalent
irreducibles, then a similar argument proves that
ComX = {©jLiQ/ci. : G C},
where di = degX^^^. Notice that the degree of X is Note also that
the dimension of ComX (as a vector space) is just k. This is because there
are k scalars Ci that can vary, whereas the identity matrices are fixed.
Next we deal with the case of sums of equivalent representations. A
convenient notation is
m
/ s
ttlX = a 0 X 0 • * • 0 X ,
where the nonnegative integer m is called the multiplicity of X.
1 . 7. COMMUTANT AND ENDOMORPHISM ALGEBRAS
25
Example 1.7.3 Suppose that
where is irreducible of degree d. Take T partitioned as before. Doing the
multiplication in TX = XT and equating blocks now yields four equations,
all of the form
for all i^j = 1,2. Corollaries 1.6.6 and 1.6.8 come into play again to reveal
that, for all i and j,
where Cij G C. Thus
^ € C fo, alH.,j (1.11)
is the commutant algebra in this case. ■
The matrices in Com2X^^^ have a name.
Definition 1.7.4 Let X = {xij) and Y be matrices. Then their tensor
product is the block matrix
X ^ F = (xijY)
/ Xi^lY xi,2F - X
3 ^ 2 , 1 ^ X2,2Y •••
V ^
Thus we could write the elements of (1.11) as
T =
Cl,l Cl, 2
C2,l C2,2
and so
ComX = {M 2 <S)Id • ^2 G Mat 2 }.
If we take X = mX^^\ then
ComX = {Mm 0 Id Mm e Mat^},
where d is the degree of X^^\ Computing degrees and dimensions, we obtain
degX = degmX^^) = mdegX^^^ = md
and
dim(ComX) = dim(M^ : Mm G Matm} =
Finally, we are led to consider the most general case:
X = © m2X(2) 0 ... 0 mkX^^\
(1.12)
26
CHAPTER 1. GROUP REPRESENTATIONS
where the are pairwise inequivalent irreducibles with degX^^^ = di. The
degree of X is given by
k
degX = ^deg(miX^"^) = ruidi -\-nri2d2 H [-rrikdk.
2=1
The reader should have no trouble combining Examples 1.7.2 and 1.7.3 to
obtain
ComX = ^ Id,) : Mm, G Mat^, for all i) (1.13)
of dimension
dim(ComX) = dim{0jLiM^. : Mm, G Matmi} = Tn\-\-m^-\ h m\.
Before continuing our investigation of the commutant algebra, we should
briefly mention the abstract vector space analogue of the tensor product.
Definition 1.7.5 Given vector spaces V and VE, then their tensor product
is the set
V CijVi (g) Wj : Cij eC,Vi e V, Wj eW}
hj
subject to the relations
(ciVi + C 2 V 2 ) (g) W = Ci(vi (g) w),+ C 2 (V 2 g) w)
and
V g (diwi + d2W2) = di(v g wi) + d2(v g W 2 ). ■
It is easy to see that V <S>W is also a vector space. In fact; the reader can
check that if S = {vi, V2, . . . , Vd] and C = {wi, W2, . . . , w/} are bases for V
and VE, respectively, then the set
{vi g Wj : I <i <d,l < j < f]
is a basis for V <^W. This gives the connection with the definition of matrix
tensor products: The algebra Mat^ has as basis the set
: l<2,j<d},
where Eij is the matrix of zeros with exactly one 1 in position (i, j). So if
X = {xij) G Matd and Y = (yk,i) G Mat/, then, by the fact that g is linear,
d f
xgy = ( Vk,iEk,i)
i.j=X kj=l
d f
2,j=l k,l=l
(1.14)
1 . 7. COMMUTANT AND ENDOMORPHISM ALGEBRAS
27
But if Eij (g) Ek.i represents the {k,l)th position of the (z, j)th block of a
matrix, then equation (1.14) says that the corresponding entry for X (g) F
should be agreeing with the matrix definition.
We return from our brief detour to consider the center of ComX. The
center of an algebra A is
Za = {a ^ A : ab — ba for all b G A].
First we will compute the center of a matrix algebra. This result should be
very reminiscent of Corollary 1.6.8 to Schur’s lemma.
Proposition 1.7.6 The center o/Mat^ is
ZMatd ~ ^ ^
Proof. Suppose that C G Then, in particular,
CEi^i = Ei^iC (1.15)
for all i. But CEi^i (respectively, Ei^iC) is all zeros except for the ith column
(respectively, row), which is the same as C’s. Thus (1.15) implies that all
off-diagonal elements of C must be 0. Similarly, if z 7^ jf, then
4" Ej,i) — {^i,j "h
where the left (respectively, right) multiplication exchanges columns (respec-
tively, rows) z and j of C. It follows that all the diagonal elements must be
equal and so C = cld for some c G C. Finally, all these matrices clearly
commute with any other matrix, so we are done. ■
Since we will be computing .^comX and the elements of the commutant
algebra involve direct sums and tensor products, we will need to know how
these operations behave under multiplication.
Lemma 1.7.7 Suppose A, X G Mat^/ and B^Y £ Mat/. Then
1 . {A®B){X®Y)=AX®BY,
2. {A (g B){X g F) = AX g BY.
Proof. Both assertions are easy to prove, so we will do only the second.
Suppose A == (uz,/) and X = (a:^,/). Then
(A g B){X g F) = {aijB)(xijY) (definition of g)
= {J2k ^i,kP ^k^y) (block multiplication)
= ( (J2k^hkXk,j) by) (distributivity)
= AX g BY. (definition of g) ■
Now consider C G .^comX, where X and ComX are given by (1.12) and
(1.13), respectively. So
CT = TC for all T G Com X.
(1.16)
28
CHAPTER 1. GROUP REPRESENTATIONS
where T = ® /dj and C = ®i=i(Cmi ® /dj. Computing the
left-hand side, we obtain
CT = (©jLiCmi ® Idi) <8i Idi) (definition of C and T)
= 0 Idi) i^mi 0 hi) (Lemma 1.7.7, item 1)
= ®'i=i{CmiMmi® hi)- (Lemma 1.7.7, item 2).
Similarly,
TC = ®ti(M„,C„, 0/dJ.
Thus equation (1.16) holds if and only if
^rrii ^rrii ~ ^rrii ^rrii f^^ ^rrii ^ .
But this just means that Crm is in the center of Mat^^ , which, by Proposition
1.7.6, is equivalent to
Crrii — ^ilrrii
for some G C. Hence
C — 02=iQ/t7i,^ 0 Idi
— ®iz=zl^ilmidi
^\Imid\ 0 ' * * 0
0 ^2/7712^2 ’ ’ ‘ ^
0 0 • • • Cklrukdk
and all members of Zqq^x have this form. Note that dimZcomX = k.
For a concrete example, let
/ 0 0 \
X=\ 0 0
V 0 0 x(2) J
where degX^^^ = 3 and degX^^^ = 4. Then the matrices T G ComX look
like
/a006000000\
0a00600000
00a0060000
cOOdOOOOOO
OcOOdOOOOO
OOcOOdOOOO ’
000000x000
0000000x00
00000000x0
\000000000x/
where a, 6, c, d, x G C. The dimension is evidently
dim(ComX) = = 2^ + 1^ = 5.
1 . 7. COMMUTANT AND ENDOMORPHISM ALGEBRAS
29
The elements C S Zcom x are even simpler:
/aOOOOOOOOOX
OnOOOOOOOO
OOaOOOOOOO
OOOaOOOOOO
OOOOaOOOOO
^ ~ OOOOOaOOOO ’
000000a:000
0000000a;00
00000000x0
yooooooooox/
where a, x S C. Here the dimension is the number of different irreducible
components of X, in this case 2.
We summarize these results in the following theorem.
Theorem 1.7.8 Let X be a matrix representation of G such that
X = © m2X^2) 0 ... 0 rrikX^^\ (1.17)
where the are inequivalent, irreducible and = di. Then
1 . deg X = ruidi rri2d2 -i [-rukdk,
2. ComX = {©•Li(Mmi © J : G Mat^^i for all i},
3. dim(ComX) = mf + H h
4. Zqqyyi X G C for all and
5. dimZcomX = k. m
What happens if we try to apply Theorem 1.7.8 to a representation Y
that is not decomposed into irreducibles? By the matrix version of Maschke’s
theorem (Corollary 1.5.4), Y is equivalent to a representation X of the form
given in equation (1.17). But if T = RXR~^ for some fixed matrix R, then
the map
T RTR-^
is an algebra isomorphism from ComX to ComT. Once the commutant
algebras are isomorphic, it is easy to see that their centers are too. Hence
Theorem 1.7.8 continues to hold with all set equalities replaced by isomor-
phisms.
There is also a module version of this result. We will use the multiplicity
notation for G-modules in the same way it was used for matrices.
Theorem 1.7.9 Let V he a G -module such that
y ^ © • • • ©
where the y^*^ are pairwise inequivalent irreducibles and dimy^'^ = d{. Then
30
CHAPTER 1. GROUP REPRESENTATIONS
1. dim V = midi + rri 2 d 2 H h rrikdk,
2. End V = Mat^. ,
3. dim(End V) = ml + + • • • + m\,
4. ^Endv isomorphic to the algebra of diagonal matrices of degree k,
and
5. dim^End V k. m
The same methods can be applied to prove the following strengthening of
Corollary 1.6.7 in the case where the field is C.
Proposition 1.7.10 Let V and W be G -modules with V irreducible. Then
dim Hom(y, W) is the multiplicity of V in W. m
1.8 Group Characters
It turns out that much of the information contained in a representation can
be distilled into one simple statistic: the traces of the corresponding matrices.
This is the beautiful theory of group characters that will occupy us for the
rest of this chapter.
Definition 1.8.1 Let X{g), ^ G G, be a matrix representation. Then the
character of X is
X{g) =ivX{g),
where tr denotes the trace of a matrix. Otherwise put, x is the map
c.
If F is a G-module, then its character is the character of a matrix represen-
tation X corresponding to V. ■
Since there are many matrix representations corresponding to a single G-
module, we should check that the module character is well-defined. But if X
and Y both correspond to V, then Y = TXT~^ for some fixed T. Thus, for
all g G G,
tvY{g)=tvTX{g)T-^ =trX{g),
since trace is invariant under conjugation. Hence X and Y have the same
character and our definition makes sense.
Much of the terminology we have developed for representations will be
applied without change to the corresponding characters. Thus if X has char-
acter X, we will say that x is irreducible whenever X is, etc. Now let us turn
to some examples.
1 .8. GROUP CHARACTERS
31
Example 1.8.2 Suppose G is arbitrary and X is a degree 1 representation.
Then the character xid) is just the sole entry of X{g) for each g E G. Such
characters are called linear characters, m
Example 1 . 8.3 Suppose we consider the defining representation of Sn with
its character If we take n = 3, then we can compute the character values
directly by taking the traces of the matrices in Example 1.2.4. The results
are
(1)(2)(3) ) = 3, (1, 2)(3) ) = 1, (1, 3)(2) ) = 1,
X‘i"f((l)(2,3)) = l, x‘“='( (1,2,3) )=0, x^"f( (1,3,2) ) = 0.
It is not hard to see that in general, if tt E 5n, then
X^^^( 7 t) = the number of ones on the diagonal of X{7r)
= the number of fixedpoints of tt. ■
Example 1 . 8.4 Let G — { 91 , 92, • • • 9n} and consider the regular repre-
sentation with module V = C[G] and character x^^®- Now X{e) = In, so
X^‘^"(e) = |G|.
To compute the character values for ^ ^ e, we will use the matrices arising
from the standard basis B = {gi, g 2 , • • • 5 gn}- Now X{g) is the permutation
matrix for the action of g on S, so is the number of fixedpoints for
that action. But if ggi = gi for any i, then we must have g — which is not
the case; i.e., there are no fixedpoints if ^ 7 ^ e. To summarize,
Y^^^(a) = / if ^
^ [0 otherwise. ■
We now prove some elementary properties of characters.
Proposition 1 . 8.5 Let X he a matrix representation of a group G of degree
d with character x-
1 - X(e) = d.
2. If K is a conjugacy class ofG, then
g,heK => x{g) = xih).
3. If Y is a representation of G with character 'if, then
x{g) = V’(fi)
for all g £ G.
Proof. 1. Since X{e) = Id,
X(e) = tr/d = d.
32
CHAPTER 1. GROUP REPRESENTATIONS
2. By hypothesis g = khk so
X{g) = tvX{g) = tvX{k)X{h)X{k)-^ = tvX{h) = x{h).
3. This assertion just says that equivalent representations have the same
character. We have already proved this in the remarks following the preceding
definition of group characters. ■
It is surprising that the converse of 3 is also true — i.e., if two represen-
tations have the same character, then they must be equivalent. This result
(which is proved as Corollary 1.9.4, part 5) is the motivation for the paragraph
with which we opened this section.
In the previous proposition, 2 says that characters are constant on conju-
gacy classes. Such functions have a special name.
Definition 1.8.6 A class function on a group G is a mapping / : G C
such that f{g) = f{h) whenever g and h are in the same conjugacy class.
The set of all class functions on G is denoted by R{G).
Clearly, the sums and scalar multiples of class functions are again class
functions, so R{G) is actually a vector space over C. Also, R{G) has a natural
basis consisting of those functions that have the value 1 on a given conjugacy
class and 0 elsewhere. Thus
dimi^(G) = number of conjugacy classes of G. (1-18)
If AT is a conjugacy class and x is a character, we can define Xk to be the
value of the given character on the given class:
Xk = Xio)
for any g e K. This brings us to the definition of the character table of a
group.
Definition 1.8.7 Let G be a group. The character table of G is an array
with rows indexed by the inequivalent irreducible characters of G and columns
indexed by the conjugacy classes. The table entry in row x and column K is
... K
XiC
By convention, the first row corresponds to the trivial character, and the first
column corresponds to the class of the identity, K = {e}. ■
It is not clear that the character table is always finite: There might be
an infinite number of irreducible characters of G. Fortunately, this turns out
1.9, INNER PRODUCTS OF CHARACTERS
33
not to be the case. In fact, we will prove in Section 1.10 that the number
of inequivalent irreducible representations of G is equal to the number of
conjugacy classes, so the character table is always square. Let us examine
some examples.
Example 1.8.8 If G = G^, the cyclic group with n elements, then each
element of Cn is in a conjugacy class by itself (as is true for any abelian
group). Since there are n conjugacy classes, there must be n inequivalent
irreducible representations of G^. But we found n degree 1 representations in
Example 1.2.3, and they are pairwise inequivalent, since they all have different
characters (Proposition 1.8.5, part 3). So we have found all the irreducibles
for Cn-
Since the representations are one-dimensional, they are equal to their
corresponding characters. Thus the table we displayed on page 5 is indeed
the complete character table for G 4 . ■
Example 1.8.9 Recall that a conjugacy class in G = <5^ consists of all per-
mutations of a given cycle type. In particular, for S 3 we have three conjugacy
classes,
Ki = {e}, 7^2 = {(1,2), (1,3), (2,3)}, and iiCa = {(1,2,3), (1,3,2)}.
Thus there are three irreducible representations of S3. We have met two of
them, the trivial and sign representations. So this is as much as we know of
the character table for S3:
Ki
K 2
Ks
1
1
1
;,( 2 )
1
-1
1
?
?
?
We will be able to fill in the last line using character inner products. ■
1.9 Inner Products of Characters
Next, we study the powerful tool of the character inner product. Taking
inner products is a simple method for determining whether a representation
is irreducible. This technique will also be used to prove that equality of
characters implies equivalence of representations and to show that the number
of irreducibles is equal to the number of conjugacy classes. First, however,
we motivate the definition.
We can think of a character x of ^ group G = {pi, ^ 2 ? • • • 5 ^n} as a row
vector of complex numbers:
X = {xi9i),x{92), ■ ■ ■ ,xi9n))-
34
CHAPTER 1. GROUP REPRESENTATIONS
If X is irreducible, then this vector can be obtained from the character table
by merely repeating the value for class K a total of |i^| times. For example,
the first two characters for in the preceding table become
= (1,1, 1,1, 1,1) and
We have the usual inner product on row vectors given by
(Ci, C2, . . . , Cn) • (^1,^2, . . . , dji) = Cidi 02^2 + * * • + Cndn^
where the bar stands for complex conjugation. Computing with our S3 char-
acters, it is easy to verify that
x(l) . ,^(1) = ^(2) . ,^,(2) ^ 0
and
x(i) • = 0.
More trials with other irreducible characters — e.g., those of C4 — will lead the
reader to conjecture that if and irreducible characters of G, then
X
(0
= { P'
if i = j,
if i ^ j.
Dividing by |G| for normality gives us one definition of the character inner
product.
Definition 1.9.1 Let x ^-nd -0 be any two functions from a group G to the
complex numbers C. The inner product of x and -0 is
g6G
Now suppose y is a G-module with character 'ip. We have seen, in the
proof of Maschke’s theorem, that there is an inner product on V itself that is
invariant under the action of G. By picking an orthonormal basis for F, we
obtain a matrix representation Y for -0, where each Y{g) is unitary; i.e.,
Y{g-^) = Y{g)-^=Y{^\
where t denotes transpose. So
ipig) =tvY{g) =tvY{g~^y = tr Y{g^^) =ifj{g~^).
Substituting this into (•, •} yields another useful form of the inner product.
Proposition 1.9.2 Let x <^'^d 0 he characters; then
ix , ^) = 07 51 x{g)i’{9~^)- ■
' ' g€G
(1.19)
1.9. INNER PRODUCTS OF CHARACTERS
35
When the field is arbitrary, equation (1.19) is taken as the definition of the
inner product. In fact, for any two functions x from G to a field, we
can define
{x, = A xig)H9~^),
geG
but over the complex numbers this “inner product” is only a bilinear form.
Of course, when restricted to characters we have • Also note
that whenever \ constant on conjugacy classes, we have
{x,i>) = -^^T.\k\xk'^k,
where the sum is over all conjugacy classes of G.
We can now prove that the irreducible characters are orthonormal with
respect to the inner product (•,•).
Theorem 1.9.3 (Character Relations of the First Kind) Let x
he irreducible characters of a group G. Then
{x,ip) =
Proof. Suppose x? '0 characters of matrix representations A, B of
degrees d, /, respectively. We will be using Schur’s lemma, and so a matrix
must be found to fulfill the role of T in Corollary 1.6.6. Let X = {xij) be a
d X f matrix of indeterminates xij and consider the matrix
( 1 . 20 )
geG
We claim that A{h)Y — YB{h) for all h e G. Indeed,
A{h)YB{h)-^ = ^ ^ A(/i)A(ff)XB(5-')5(/i-i)
geG
geG
= jkY Mg)XB{g-^)
' ' geG
g=hg
= Y,
and our assertion is proved. Thus by Corollaries 1.6.6 and 1.6.8,
y =
0 if A ^ B,
cld if A = jB.
( 1 . 21 )
36
CHAPTER 1. GROUP REPRESENTATIONS
Consider first the case where % 7^ so that A and B must be inequivalent.
Since this forces yi^j = 0 for every element of F, we can take the (i, j) entry
of equation (1.20) to obtain
1
EE
k,l geG
for all z, j. If this polynomial is to be zero, the coefficient of each Xk,i must
also be zero, so
geG
for all z, j, /c, 1. Notice that this last equation can be more simply stated as
( 1 . 22 )
since our definition of inner product applies to all functions from G to C.
Now,
and
so
X = trA = ai,i -f tt2,2 H h ad,d
ip = tv B = 61,1 + 62,2 H h bfj,
(x,V^) == {x,'tpy = Ey («*.»> = 0
as desired.
Now suppose x = Since we are only interested in the character values,
we might as well take A — B also. By equation (1.21), there is a scalar
c G C such that yi^j = cdij. So, as in the previous paragraph, we have
j)' == 0 as long as z 7^ j. To take care of the case z = j, consider
A\'E^^9)XA{g ^) = cld
' ' geG
and take the trace on both sides:
cd
trcid
-rpr^tTA{g)XA(g-^)
' ' geG
1
geG
trX.
Thus
= c — 2 trX, which can be rewritten as
1
EE ai,k{ 9 )xk,iai,iig = ^(^1.1 + ^2,2 H + Xd,d)-
k,l geG
1.9. INNER PRODUCTS OF CHARACTERS
37
Equating coefficients of like monomials in this equation yields
1
W\
'^ai^k{g)ai,i{g
geG
(1.23)
It follows that
ix,x) =
d
i=l
d
= El
i=l
1 ,
and the theorem is proved. ■
Note that equations (1.22) and (1.23) give orthogonality relations for the
matrix entries of the representations.
The character relations of the first kind have many interesting conse-
quences.
Corollary 1.9.4 Let X be a matrix representation of G with character x-
Suppose
X S TniX(i) © © ... © ,
where the are pairwise inequivalent irreducibles with characters x^’’Y
1. X = + ni2X^^^ H H mkX^'='>.
2. (X)X^'’^) = "ij for all j.
3. ix,x) = m? + m| H +m|.
4. X is irreducible if and only if (x, x) = 1*
5. Let Y be another matrix representation of G with character xp. Then
X if and only if x{g) =
for all g £ G.
Proof. 1. Using the fact that the trace of a direct sum is the sum of the
traces, we see that
k k
^z=ztrX = tr ^rUiX^''^ =
i=l 2=1
38
CHAPTER 1. GROUP REPRESENTATIONS
2. We have, by the previous theorem,
i i
3. By another application of Theorem 1.9.3:
ix,x) = = Yl'^i
i j i,j i
4. The assertion that X is irreducible implies that (x^x) = 1 is just part
of the orthogonality relations already proved. For the converse, suppose that
(x,x)
i
Then there must be exactly one index j such that mj = 1 and all the rest of
the TTii must be zero. But then X = X^^\ which is irreducible by assumption.
5. The forward implication was proved as part 3 of Proposition 1.8.5. For
the other direction, let Y = There is no harm in assuming that
the X and Y expansions both contain the same irreducibles: Any irreducible
found in one but not the other can be inserted with multiplicity 0. Now
X = so (x?X^*^) = ('05 X^^^) foi* But then, by part 2 of this corollary,
nrii = rii for all i. Thus the two direct sums are equivalent — i.e., X = Y. m
As an example of how these results are applied in practice, we return
to the defining representation of <Sn- To simplify matters, note that both
7r,7r“^ G Sn have the same cycle type and are thus in the same conjugacy
class. So if X is a character of 5n, then xM = x (^~^)5 since characters are
constant on conjugacy classes. It follows that the inner product formula for
Sn can be rewritten as
ix,i>) = ^ Y xWV'W- (1-24)
* TCeSn
Example 1.9.5 Let G = S 3 and consider x = X^^^- Let Le
the three irreducible characters of <S 3 , where the first two are the trivial and
sign characters, respectively. By Maschke’s theorem, we know that
X = + ^2X^^^ d-
Furthermore, we can use equation (1.24) and part 2 of Corollary 1.9.4 to com-
pute mi and m 2 (character values for x = X^^^ were found in Example 1.8.3):
mi = ^ E = ^(3-1 + M + M + M+O-l+O-l) = 1,
* -n-e<S3
m 2 = ix,X^^^) ^ E = i(3-l-M-l-l-M+0-l+0-l) = 0.
TrGtSa
1.9. INNER PRODUCTS OF CHARACTERS
39
Thus
X = X^^^ +m3X^^^.
In fact, we already knew that the defining character contained a copy
of the trivial one. This was noted when we decomposed the corresponding
matrices as X = A^B, where A was the matrix of the trivial representation
(see page 13). The exciting news is that the B matrices correspond to one or
more copies of the mystery character These matrices turned out to be
5(e)=(j J), 5((1,2))=(-J
B((l,3)) = _} _5 B((2,3)) J
S( (1,2,3) )=( “J “J), S( (1,3,2) ) _iy
If we let t/; be the corresponding character, then
i>{e) = 2,
(1,2)) =V>( (1,3))=0( (2,3) )=0,
V-( (1,2,3)) (1,3,2)) = -!.
If Ip is irreducible, then m3 = 1 and we have found (If iiot) being
of degree 2, must contain two copies of x^^^-) But part 4 of Corollary 1.9.4
makes it easy to determine irreducibility; merely compute:
{xp, ip) = + 0^ + + 0^ + (-1)2 + (-1)2) = 1.
0
We have found the missing irreducible. The complete character table for Ss
is thus
Ki
K2
Ks
1
1
1
x(2)
1
-1
1
x(3)
2
0
-1
In general, the defining module for 5n, V = C{1, 2 , . . . , n}, always has
W = C{1 + 2 -f- • • • + n} as a submodule. If x^^^ and X“^ are the characters
corresponding to W and respectively, then V = W^W-^. This translates
to
X<lef ,,, ^(1)
on the character level. We already know that x^^^ counts fixedpoints and
that x^^^ is the trivial character. Thus
= (number of fixedpoints of tt) — 1
is also a character of Sn- In fact, x~^ is irreducible, although that is not
obvious from the previous discussion. ■
40
CHAPTER 1. GROUP REPRESENTATIONS
1.10 Decomposition of the Group Algebra
We now apply the machinery we have developed to the problem of decom-
posing the group algebra into irreducibles. In the process, we determine the
number of inequivalent irreducible representations of any group.
Let G be a group with group algebra C[G] and character x =
Maschke’s theorem (Theorem 1.5.3), we can write
C[G]-0m,VW, (1.25)
i
where the run over all pairwise inequivalent irreducibles (and only a finite
number of the rrii are nonzero).
What are the multiplicities rrii? If has character x^*^ then, by part
2 of Corollary 1.9.4,
rrii =
see
But we computed the character of the regular representation in Example 1.8.4,
and it vanished for ^ ^ e with x(e) = |G|. Plugging in the preceding values,
we obtain
mi = |^x(e)x^‘^(e) = (1.26)
by Proposition 1.8.5, part 1. Hence every irreducible G- module occurs in
C[G] with multiplicity equal to its dimension. In particular, they all appear
at least once, so the list of inequivalent irreducibles must be finite (since the
group algebra has finite dimension). We record these results, among others
about the decomposition of C[G], as follows.
Proposition 1.10.1 Let G be a finite group and suppose C[G] =
where the form a complete list of pairwise inequivalent irreducible G-
modules. Then
1. mi = dim
2. = \G\, and
3. The number ofV^'^^ equals the number of conjugacy classes of G.
Proof. Part 1 is proved above, and from it, part 2 follows by taking dimen-
sions in equation (1.25).
For part 3, recall from Theorem 1.7.9 that
number of = dim^EndCfC]-
What do the elements of EndC[G] look like? Given any v G C[G], define the
map : C[G] -)> C[G] to be right multiplication by v, i.e..
0v(w) = wv.
1.10. DECOMPOSITION OF THE GROUP ALGEBRA
41
for all w G C[G]. It is easy to verify that G EndC[G]. In fact, these are
the only elements of EndC[G] because we claim that C[G] = EndC[G] as
vector spaces. To see this, consider (p : C[G] -> EndC[G] given by
V 0,.
Proving that 0 is linear is not hard. To show injectivity, we compute its
kernel. If 0^ is the zero map, then
0 = 0^(e) = €v = V.
For surjectivity, suppose 6 G EndC[G] and consider ^(e), which is some
vector V. It follows that 0 = because given any g ^ G,
^(g) = = gd(e) = gv = gv = <t>^(g),
and two linear transformations that agree on a basis agree everywhere. On
the algebra level, our map 0 is an anti-isomorphism, since it reverses the
order of multiplication: 0^0^ = 0^^ for all v, w G C[G]. Thus 0 induces an
anti-isomorphism of the centers of C[G] and EndC[G], so that
number of = dimZ^fG]-
To find out what the center of the group algebra looks like, consider any
z = cigi + • • • + Cngn ^ -^c[G]? where the gi are in G. Now for all G G, we
have zh = hz, or z = hzh“^, which can be written out as
Clgl H h Cngn = Clhgih~^ H h c„hg„h~^
But as h takes on all possible values in G, hg\h~^ runs over the conjugacy
class of gi. Since z remains invariant, all members of this class must have the
same scalar coefficient ci. Thus if G has k conjugacy classes Ki, . . . , and
we let
for i == 1, . . . , A;, then we have shown that any z G Zc[g] can be written as
k
z = ^ ^ .
2 = 1
Similar considerations show that the converse holds: Any linear combination
of the Zi is in the center of C[G]. Finally, we note that the set {zi, . . . , Zk}
forms a basis for Zc[g]- We have already shown that they span. They must
also be linearly independent, since they are sums over pairwise disjoint subsets
of the basis {g : g £ G] oi C[G]. Hence
number of conjugacy classes = dim Zc[g] number of 1/^^^
as desired. ■
As a first application of this proposition, we derive a slightly deeper rela-
tionship between the characters and class functions.
42
CHAPTER 1. GROUP REPRESENTATIONS
Proposition 1.10.2 The irreducible characters of a group G form an or-
thonormal basis for the space of class functions R{G).
Proof. Since the irreducible characters are orthonormal with respect to the
bilinear form (*,•) on R{G) (Theorem 1.9.3), they are linearly independent.
But part 3 of Proposition 1.10.1 and equation (1.18) show that we have
dimi?(G) such characters. Thus they are a basis. ■
Knowing that the character table is square permits us to derive orthogo-
nality relations for its columns as a companion to those for the rows.
Theorem 1.10.3 (Character Relations of the Second Kind) LetK^L
be conjugacy classes of G. Then
E
— 1^1 A
XkXl =
where the sum is over all irreducible characters of G.
Proof. If X -0 are irreducible characters, then the character relations of
the first kind yield
(Xi V') = E ^
where the sum is over all conjugacy classes of G. But this says that the
modified character table
U = [^/\K\JW\ Xk)
has orthonormal rows. Hence [/, being square, is a unitary matrix and has
orthonormal columns. The theorem follows. ■
As a third application of these ideas, we can now give an alternative
method for finding the third line of the character table for that does not
involve actually producing the corresponding representation. Let the three
irreducible characters be x^^\ where the first two are the
trivial and sign characters, respectively. If d denotes the dimension of the
corresponding module for then by Proposition 1.10.1, part 2,
12 + 1 ^ + ^2 = 1531 = 6 .
Thus — d=2. To find the value of on any other permutation, we
use the orthogonality relations of the second kind. For example, to compute
:r-x("H(l,2) ),
0 = E X^‘^(e)x(*K(l,2)) = 1 • 1 + 1(-1) + 2af,
SO X = 0.
1.11. TENSOR PROD UCTS AGAIN
43
1.11 Tensor Products Again
Suppose we have representations of groups G and H and wish to construct
a representation of the product group G x H. It turns out that we can use
the tensor product introduced in Section 1.7 for this purpose. In fact, all the
irreducible representations of GxH can be realized as tensor products of irre-
ducibles for the individual groups. Proving this provides another application
of the theory of characters.
Definition 1.11.1 Let G and H have matrix representations X and F,
respectively. The tensor product representation^ X ^ Y, assigns to each
(g^h) ^ G X H the matrix
{X^Y){g,h) = X{g)^Y{h).m
We must verify that this is indeed a representation. While we are at it, we
might as well compute its character.
Theorem 1.11.2 Let X and Y be matrix representations for G and H, re-
spectively.
1. Then X ^Y is a representation of G x H.
2. If X,Y andX^Y have characters denoted by x, andx^i^i respec-
tively, then
= x{.g)'<P{h)
for all (g, h) £ G x H.
Proof. 1. We verify the two conditions defining a representation. First of
all,
{X <S> Y){e, e) = X{e) ® Y{e) = J ® / = J.
Secondly, if {g,h), {g',h') £ G x H, then using Lemma 1.7.7, part 2,
{X0Y)iig,h)-ig',h')) = {X®Y){gg',hh')
= X{gg')®Y{hh')
- X{g)X{g')®Y{h)Y{h')
= {X{g)®Y{h))-{X{g')®Y{h'))
= {X®Y){g,h)-{X®Y){g',h').
2. Note that for any matrices A and B,
trA®B = tr{aijB) = ^ trB = ivAtrB.
Thus
iX'S)ip)ig,h) =tr(X(g)®Y{h)) = tTX{g)trY{h) =x{g)i>{h)-
44
CHAPTER!. GROUP REPRESENTATIONS
There is a module-theoretic way to view the tensor product of represen-
tations. Let F be a G-module and W be an i7-module. Then we can turn
the vector space V <S>W into a G x if-module by defining
(p, h)(v 0 w) = {gv) 0 (hw)
and linearly extending the action as v and w run through bases of V and W,
respectively. It is not hard to show that this definition satisfies the module
axioms and is independent of the choice of bases. Furthermore, if V and W
correspond to matrix representations X and Y via the basis vectors consisting
of v’s and w’s, then F 0 IF is a module for X 0 F in the basis composed of
V 0 w’s.
Now we show how the irreducible representations of G and H completely
determine those of G x H.
Theorem 1.11.3 Let G and H he groups.
1. If X and Y are irreducible representations of G and H, respectively,
then X <S)Y is an irreducible representation of G x H.
2. If X^'^^ and are complete lists of inequivalent irreducible represen-
tations for G and H, respectively, then X^'^^ 0 Y^^^ is a complete list of
inequivalent irreducible G x H -modules.
Proof. 1. If 0 is any character, then we know (Corollary 1.9.4, part 4)
that the corresponding representation is irreducible if and only if (0,0) = 1.
Letting X and F have characters x 0? respectively, we have
(X<^ 05X^0)
F ix®i’){9,h){x®'>P){9 \h
I I u
(g,h)SGxH
1*^1 9€G
= (x.x)(^.V’)
= 1-1
= 1 .
geG heH
2. Let and Y^^^ have characters and 0^-^\ respectively. Then as
in the proof of 1, we can show that
Thus from Corollary 1.9.4, part 3, we see that the ^ 0^"^^ are pairwise
inequivalent.
To prove that the list is complete, it suffices to show that the number of
such representations is the number of conjugacy classes of G x H (Proposi-
tion 1.10.1, part 3). But this is the number of conjugacy classes of G times
the number of conjugacy classes of H, which is the number of X^^^ 0 Y^^\ m
1.12. RESTRICTED AND INDUCED REPRESENTATIONS
45
1.12 Restricted and Induced Representations
Given a group G with a subgroup is there a way to get representations
of G from those of H or vice versa? We can answer these questions in the
affirmative using the operations of restriction and induction. In fact, we have
already seen an example of the latter, namely, the coset representation of
Example 1.3.5.
Definition 1.12.1 Consider H < G and a matrix representation X of G.
The restriction of X to if, is given by
(h) = X{h)
for all h £ H. If has character x> then denote the character of by
xIh- ■
It is trivial to verify that X\.^ is actually a representation of H. If the group
G is clear from context, we will drop it in the notation and merely write X\,h>
Note that even though X may be an irreducible representation of G, X]^h
need not be irreducible on H.
The process of moving from a representation of H to one of G is a bit more
involved. This construction, called induction, is due to Frobenius. Suppose
y is a matrix representation of if. We might try to obtain a representation of
G by merely defining Y to be zero outside of if. This will not work because
singular matrices are not allowed. However, there is a way to remedy the
situation.
Definition 1.12.2 Consider if < G and fix a transversal ti, . . . ,t/ for the
left cosets of if, i.e., G = t\H (+)•••(+) UH, where |+) denotes disjoint union.
If y is a representation of fi, then the corresponding induced representation
yf^ assigns to each g £ G the block matrix
ng {g) = {Y{t-^gtj)) =
( y{h ^gh)
Y{t^^gh)
Y{tT^gt2)
Y{q^gt2)
\ Y(ti ^gti) Y{ti ^gt2)
where Y (g) is the zero matrix ii g ^ H. m
YitT^gti) \
Yit^^gti)
Yitr^gti) /
We will abbreviate y to y if no confusion will result; this notation
applies to the corresponding characters as well. It is not obvious that Y^h
is actually a representation of G, but we will postpone that verification until
after we have looked at an example.
As usual, let G = Ss and consider H — {e, (2,3)} with the transversal
G = ff y (l,2)fr(+) (l,3)ff as in Example 1.6.3. Let y == 1 be the trivial
representation of H and consider X = It^. Calculating the first row of the
46
CHAPTER 1. GROUP REPRESENTATIONS
matrix for the transposition (1,2) yields
Y{ 2)r ) = Y{ (1, 2) ) = 0, (since (1, 2) 0 H)
Y{ e-i(l, 2)(1, 2) ) = F( e ) = 1, (since e e H)
y(e-i(l,2)(l,3) ) = r( (1,3,2) )=0. (since (1, 3, 2) ^ F)
Continuing in this way, we obtain
/O 1 0\
X{ (1,2) )= 10 0.
V 0 0 1 /
The matrices for the coset representation are beginning to appear again. This
is not an accident, as the next proposition shows.
Proposition 1.12.3 Let H < G have transversal {ti,...,ti} with cosets
T-L = {tiH^ . . . ^tiH}. Then the matrices 0 / 1 are identical with those
of G acting on the basis TL for the coset module VTL.
Proof. Let the matrices for and be X = and Z =
respectively. Both arrays contain only zeros and ones. Finally, for any g £ G^
= 1 <=> e H
gtjH = UH
^ = 1 - ■
Thus CH is a module for If^-
It is high time that we verified that induced representations are well-
defined.
Theorem 1.12.4 Suppose H < G has transversal {ti, . . . ,t/} and let Y he a
matrix representation of H. Then X = is a representation of G.
Proof. Analogous to the case where Y is the trivial representation, we prove
that X{g) is always a block permutation matrix; i.e., every row and column
contains exactly one nonzero block Y{t~^gtj). Consider the first column (the
other cases being similar). It suffices to show that there is a unique element
of H on the list tf^gti^t^^gti^ . . . But gt\ G UH for exactly one of
the U in our transversal, and so t~^gt\ £ H is the element we seek.
We must first verify that X{e) is the identity matrix, but that follows
directly from the definition of induction.
Also, we need to show that X{g)X{h) = X{gh) for all g,h £ G, Consid-
ering the (i, j) block on both sides, it suffices to prove
Y^Y{t-^gtk)Y{tl^htj) = Y{t-^ghtj).
k
For ease of notation, let ak = tf^gtk, bk = and c = t~^ghtj. Note
that akbk = c for all k and that the sum can be rewritten
^y(afc)y(6fe)^y(c).
k
1.12. RESTRICTED AND INDUCED REPRESENTATIONS
47
Now the proof breaks into two cases.
If Y (c) — 0, then c ^ and so either at ^ H ov bk ^ H for all k. Thus
Y{ak) or Y{bk) is zero for each /c, which forces the sum to be zero as well.
If Y (c) ^ 0, then c ^ H. Let m be the unique index such that am G H.
Thus bm = G if, and so
Y,y{ak)Y{bk) = Y{am)Y{bm) = Y{aM = Y(c),
k
completing the proof. ■
It should be noted that induction, like restriction, does not preserve irre-
ducibility. It might also seem that an induced representation depends on the
transversal and not just on the subgroup chosen. However, this is an illusion.
Proposition 1.12.5 Consider H < G and a matrix representation Y of
H . Let {t\^. . . ^ti] and {si, . . . , s/} be two transversals for H giving rise to
representation matrices X and Z, respectively, for Y^^ . Then X and Z are
equivalent.
Proof. Let x, '0, and 0 be the characters of X, T, and Z, respectively. Then
it suffices to show that x — (Corollary 1.9.4, part 5). Now
xio) = (1-27)
i i
where 'ip(g) = 0 i^ g ^ H . Similarly,
i
Since the U and Si are both transversals, we can permute subscripts if nec-
essary to obtain UH = siH for all i. Now ti = sihi, where hi £ H for all i,
and so
t~^gti = hf^s'^gsihi.
Thus t~^gti G if and only if s~^gsi G H, and when both lie in H, they are
in the same conjugacy class. It follows that 'ip{t~^gti) = ^|J{sf^gsi), since -0
is constant on conjugacy classes of H and zero outside. Hence the sums for
X and 0 are the same. ■
We next derive a useful formula for the character of an induced repre-
sentation. Let H, 0, and the U be as in the preceding proposition. Then
^{t^^gti) = 'ip{h~^t^^gtih) for any h e H, so equation (1.27) can be rewrit-
ten as
( 5 ) = 07 XI XI
' I i heH
But as h runs over H and the U run over the transversal, the product tih
runs over all the elements of G exactly once. Thus we arrive at the identity
(5) = 707 X
x^G
(1.28)
48
CHAPTER!. GROUP REPRESENTATIONS
This formula will permit us to prove the celebrated reciprocity law of
Probenius, which relates inner products of restricted and induced characters.
Theorem 1.12.6 (Probenius Reciprocity) Let H < G and suppose that
'0 and X characters of H and G, respectively. Then
= (V'.X'Ih),
where the left inner product is calculated in G and the right one in H.
Proof. We have the following string of equalities:
W^,x) = X] (ff)x(5
geG
= ickSE ip{x ^gx)x{g
' " ' xeGgeG
fp{y)x{xy
' " ' xeGyeG
= jGPi E E i>(y)x(v~')
x^G yEG
' ' yeG
= H V’(y)x(y“^)
' ' veH
= (V’,x4-if>- ■
(equation (1.28))
(let y = x~^gx)
(x constant on G’s classes)
{x constant in the sum)
(0 is zero outside H)
1.13 Exercises
1. An inversion in tt == xi, X 2 , . . . , G Sn (one-line notation) is a pair
Xi , Xj such that i < j and Xi > Xj . Let inv tt be the number of inversions
of TT.
(a) Show that if tt can be written as a product of k transpositions,
then k = invTr (mod 2).
(b) Use part (a) to show that the sign of tt is well-defined.
2. If group G acts on a set S and s e S, then the stabilizer of s is Gs =
{geG : gs = 5 }. The orbit of s is Os = [gs : g e G}.
(a) Prove that Gs is a subgroup of G.
(b) Find a bijection between cosets of G/Gs and elements of Os-
(c) Show that \Os\ = |G|/|Gs| and use this to derive formula (1.1) for
\Kg\.
3. Let G act on S with corresponding permutation representation CS.
Prove the following.
1.13. EXERCISES
49
(a) The matrices for the action of G in the standard basis are permu-
tation matrices.
(b) If the character of this representation is x g G G, then
x{g) = the number of fixed points of g acting on 5.
4. Let G be an abelian group. Find all inequivalent irreducible represen-
tations of G. Hint: Use the fundamental theorem of abelian groups.
5. If X is a matrix representation of a group G, then its kernel is the
set N = {g ^ G : X{g) — I}. A representation is faithful if it is
one-to-one.
(a) Show that X is a normal subgroup of G and find a condition on
N equivalent to the representation being faithful.
(b) Suppose X has character x and degree d. Prove that g E N il and
only if x{g) = d- Hint: Show that xid) is a sum of roots of unity.
(c) Show that for the coset representation, N = {ligiHgf^, where the
gi are the transversal.
(d) For each of the following representations, under what conditions
are they faithful: trivial, regular, coset, sign for 5n, defining for
5n, degree 1 for Gn?
(e) Define a function Y on the group G/N by Y{gN) — X{g) for
gN e G/N.
i. Prove that F is a well-defined faithful representation of G/N.
ii. Show that Y is irreducible if and only if X is.
iii. If X is the coset representation for a normal subgroup H of
G, what is the corresponding representation Y?
6. It is possible to reverse the process of part (e) in the previous exercise.
Let N be any normal subgroup of G and let F be a representation of
G/N. Define a function on G by X{g) = Y{gN).
(a) Prove that X is a representation of G. We say that X has been
lifted from the representation F of G/N.
(b) Show that if F is faithful, then X has kernel N.
(c) Show that X is irreducible if and only if F is.
7. Let X be a reducible matrix representation with block form given by
equation (1.4). Let F be a module for X with submodule W correspond-
ing to A. Consider the quotient vector space V/W = {v-fIF : vgF}.
Show that V/W is a G-module with corresponding matrices C{g). Fur-
thermore, show that we have
v^we{v/w).
50
CHAPTER 1. GROUP REPRESENTATIONS
8. Let F be a vector space. Show that the following properties will hold
in V if and only if a related property holds for a basis of V.
(a) F is a G-module.
(b) The map 0 : F -> VF is a G- homomorphism.
(c) The inner product (*, •) on F is G-invariant.
9. Why won’t replacing the inner products in the proof of Maschke’s the-
orem by bilinear forms give a demonstration valid over any field? Give
a correct proof over an arbitrary field as follows. Assume that A is a
reducible matrix representation of the form (1.4).
(a) Write out the equations obtained by equating blocks in X{gh) =
X{g)X{h). What interpretation can you give to the equations
obtained from the upper left and lower right blocks?
(b) Use part (a) to show that
TX{g)T-^
Mg) 0 \
0 C{g) )
where T =
D
and D
1
|c?l
Z,eGM9-^)B{g).
10. Verify that the map X : GL 2 given in the example at the end of
Section 1.5 is a representation and that the subspace W is invariant.
11. Find H <Sn and a set of tabloids 5 such that CH = CS = C{1, 2, . . . , n}.
12. Let X be an irreducible matrix representation of G. Show that \ig ^ Zq
(the center of G), then X{g) = cl for some scalar c.
13. Let {Ai, A 2 , . . . , An} C GLd be a subgroup of commuting matrices.
Show that these matrices are simultaneously diagonalizable using rep-
resentation theory.
14. Prove the following converse of Schur’s lemma. Let A be a represen-
tation of G over C with the property that only scalar multiples cl
commute with X{g) for all g E G. Prove that A is irreducible.
15. Let A and Y be representations of G. The inner tensor product^ X^Y,
assigns to each g £ G the matrix
{X®Y)ig) = X{g)0Y{g).
(a) Verify that A^F is a representation of G.
(b) Show that if A, F, and A(g)F have characters denoted by x?
and X ‘^'05 respectively, then {x^'^){q) — x(5')'0(p)-
(c) Find a group with irreducible representations A and F such that
X®Y is not irreducible.
1.13. EXERCISES
51
(d) However, prove that if X is of degree 1 and Y is irreducible, then
so is X(§Y.
16. Construct the character table of 54 . You may find the lifting process of
Exercise 6 and the inner tensor products of Exercise 15 helpful.
17. Let Dn be the group of symmetries (rotations and refiections) of a
regular n-gon. This group is called a dihedral group.
(a) Show that the abstract group with generators p, r subject to the
relations
p'^ = = e and pr = rp~^
is isomorphic to Dn.
(b) Conclude that every element of Dn is uniquely expressible as r*p^,
where 0 < i < 1 and 0 < j < n — 1.
(c) Find the conjugacy classes of Dn-
(d) Find all the inequivalent irreducible representations of Dn- Hint:
Use the fact that Cn is a normal subgroup of Dn-
18. Show that induction is transitive as follows. Suppose we have groups
G > H > K and a matrix representation X of K. Then
Chapter 2
Representations of the
Symmetric Group
In this chapter we construct all the irreducible representations of the sym-
metric group. We know that the number of such representations is equal to
the number of conjugacy classes (Proposition 1.10.1), which in the case of Sn
is the number of partitions of n. It may not be obvious how to associate an
irreducible with each partition A = (Ai, A 2 , . . . , A/), but it is easy to find a
corresponding subgroup S\ that is an isomorphic copy of <Sai x 5^2 x • • • x S\i
inside We can now produce the right number of representations by in-
ducing the trivial representation on each S\ up to Sn-
If is a module for It^^? then it is too much to expect that these
modules will all be irreducible. However, we will be able to find an ordering
A(i),A^^\... of all partitions of n with the following nice property. The
first module will be irreducible; call it Next, will contain
only copies of plus a single copy of a new irreducible In general,
will decompose into some for k < i and a unique new irreducible
called the ith Specht module. Thus the matrix giving the multiplicities
for expressing as a direct sum of the will be lower triangular
with ones down the diagonal. This immediately makes it easy to compute
the irreducible characters of Sn- We can also explicitly describe the Specht
modules themselves.
Much of the material in this chapter can be found in the monograph of
James [Jam 78], where the reader will find a more extensive treatment.
2.1 Young Subgroups, Tableaux, and Tabloids
Our objective in this section is to build the modules M^. First, however, we
will introduce some notation and definitions for partitions.
If A = (Ai, A 2 , . . . , A/) is a partition of n, then we write A h n. We also
53
54 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
use the notation |A| = ^ partition of n satisfies |A| = n. We
can visualize A as follows.
Definition 2.1.1 Suppose A = (Ai, A 2 , . . . , A/) h n. The Ferrers diagram,
or shape, of A is an array of n dots having I left-justified rows with row i
containing Xi dots for 1 < i < /. ■
The dot in row i and column j has coordinates (i,j), as in a matrix.
Boxes (also called cells ) are often used in place of dots. As an example, the
partition A = (3, 3, 2 , 1 ) has Ferrers diagram
• • •
• • •
or
• •
•
where the box in the (2, 3) position has an X in it. The reader should be aware
that certain authors, notably Francophones, write their Ferrers diagrams as
if they were in a Cartesian coordinate system, with the first row along the
x-axis, the next along the line y = I, etc. With this convention, our example
partition would be drawn as
•
• •
• • •
• • •
but we will stick to “English” notation in this book. Also, we will use the
symbol A to stand for both the partition and its shape.
Now we wish to associate with A a subgroup of Sn- Recall that if T is any
set, then St is the set of permutations of T.
Definition 2.1.2 Let A = (Ai, A 2 , . . . , A^) h n. Then the corresponding
Young subgroup of Sn is
S\ = <S{i,2,...,Ai} X <5{Ai + 1,Ai+2,...,Ai+A 2} X • • • X 5{^_Ai + l,n-Ai+2,...,n} • ■
These subgroups are named in honor of the Reverend Alfred Young, who was
among the first to construct the irreducible representations of Sn [You 27,
You 29]. For example,
*5(3, 3 , 2,1) = ^{1,2,3} X X <S(7^8} X 5{g}
^ ^3 X ^3 X ^2 X 5i,
In general, 5 (Ai,a 2 ,...,ao and 5 ai x 5 a 2 x • • • x Sx^ are isomorphic as groups.
Now consider the representation • If tti , 7 T 2 , . . . , tt^ is a transversal for
S\, then by Proposition 1.12.3 the vector space
= C{7Ti5a, 7T25a, . . . , TTfe^A}
is a module for our induced representation. The cosets can be thought of
geometrically as certain arrays in the following way.
2.L YOUNG SUBGROUPS, TABLEAUX, AND TABLOIDS
55
Definition 2.1.3 Suppose A h n. A Young tableau of shape A, is an array t
obtained by replacing the dots of the Ferrers diagram of A with the numbers
1, 2, . . . , n bijectively. ■
Let ti^j stand for the entry of t in position A Young tableau of shape A
is also called a X-tableau and denoted by Alternatively, we write sht = X.
Clearly, there are nl Young tableaux for any shape A h n. To illustrate, if
A = (2,l)= • • ,
then a list of all possible tableaux of shape A is
12 21 13 31 23 32
3 ’ 3 ’ 2 ’ 2 ’ 1 ’ 1 '
The first tableau above has entries = 1, = 2, and ^ 2 ,i == 3.
Young tabloids were mentioned in Example 1.6.3. We can now introduce
them formally as equivalence classes of tableaux.
Definition 2.1.4 Two A-tableaux ti and t 2 are row equivalent, t\ ~ ^ 2 , if
corresponding rows of the two tableaux contain the same elements. A tabloid
of shape A, or X-tabloid, is then
{i} = {ti\ti ~ t}
where sht = X. m
As before, we will use lines between the rows of an array to indicate that it
is a tabloid. So if
1 2
3
2 2 1 \ _ 1 2
’3 J ~ A '
If A = (Ai, A 2 , . . . , A/) h n, then the number of tableaux in any given equiv-
alence class is Ai!A 2 !---A/! A!. Thus the number of A-tabloids is just
n!/A!.
Now 7T e Sn acts on a tableau t = (Uj) of shape A h n as follows:
TTt = (Tr(tij))
For example,
(1,2,3).J 2 J L
This induces an action on tabloids by letting
■K{t) = {ttO-
The reader should check that this is well defined — i.e., independent of the
choice of t. Finally, this action gives rise, in the usual way, to an <Sn-module.
then
W = {3
56 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
Definition 2.1.5 Suppose A h n. Let
where {ti}, . . . , {tk} is a complete list of A-tabloids. Then is called the
permutation module corresponding to X. u
Since we are considering only row equivalence classes, we could list the rows of
in any order and produce an isomorphic module. Thus is defined for
any composition (ordered partition) (i. As will be seen in the next examples,
we have already met three of these modules.
Example 2.1.6 If A = (n), then
= C{ 12 ••• n }
with the trivial action. ■
Example 2.1.7 Now consider A = Each equivalence class {t} consists
of a single tableau, and this tableau can be identified with a permutation in
one-line notation (by taking the transpose, if you wish). Since the action of
Sn is preserved,
^ CSn,
and the regular representation presents itself. ■
Example 2.1.8 Finally, if A = (n — 1,1), then each A-tabloid is uniquely
determined by the element in its second row, which is a number from 1 to n.
As in Example 1.6.3, this sets up a module isomorphism
^C{l,2,...,n},
SO we have recovered the defining representation. ■
Example 2.1.9 Let us compute the full set of characters of these modules
when n = 3. In this case the only partitions are A = (3), (2, 1), and (1^). From
the previous examples, the modules correspond to the trivial, defining,
and regular representations of S3, respectively. These character values have
all been computed in Chapter 1. Denote the character of by and
represent the conjugacy class of S3 corresponding to by Then we have
the following table of characters:
2.2. DOMINANCE AND LEXICOGRAPHIC ORDERING
57
The enjoy the following general property of modules.
Definition 2.1.10 Any G-module M is cyclic if there is a v € M such that
M - CGv,
where Gv = G G}. In this case we say that M is generated by v. ■
Since any A-tabloid can be taken to any other tabloid of the same shape by
some permutation, is cyclic. We summarize in the following proposition.
Proposition 2.1.11 If X \- n, then is cyclic, generated by any given
X-tabloid. In addition, dimM^ = n\/X\, the number of X-tabloids. m
What is the connection between our permutation modules and those ob-
tained by inducing up from a Young subgroup? One can think of
<5{Ai + 1,Ai+2,..
..,Ai+A2} ^
X 5|7T,_Aj-f-l,n— Ai+2,...,n}
as being modeled by the tabloid
1
2
• Ai
^ Ai + 1
r J.A 1
Ai + 2
• Ai + A 2
n — A/ + 1 • • • n
The fact that 5 a contains all permutations of a given interval of integers is
mirrored by the fact that the order of these integers is immaterial in t^ (since
they all occur in the same row). Thus the coset 7 t5a corresponds in some way
to the tabloid {7rt^}. To be more precise, consider Theorem 2.1.12.
Theorem 2.1.12 Consider A h n with Young subgroup S\ and tabloid
as before. Then = C5n5A and = CSn{t^} nre isomorphic as Sn~
modules.
Proof. Let tti , 7T2 , . . . , be a transversal for 5 a . Define a map
by for 2 = 1, . . . , A; and linear extension. It is not hard to
verify that 6 is the desired 5n -isomorphism of modules. ■
2.2 Dominance and Lexicographic Ordering
We need to find an ordering of partitions A such that the have the nice
property of the introduction to this chapter. In fact, we consider two impor-
tant orderings on partitions of n, one of which is only a partial order.
58 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
Definition 2.2.1 If A is a set, then a partial order on A is a relation < such
that
1. a < a,
2. a <b and b < a implies a = 6, and
3. a <b and b < c implies a < c
for all a,b,c e A. In this case we say that {A, <) is a partially ordered set^ or
poset for short. We also write 5 > a for a < & as well as a < 6 (or b > a) for
a <b and a ^ b.
If, in addition, for every pair of elements a^b e A we have either a < b or
b < a^ then < is a total order and (^, <) is a totally ordered set. m
The three laws for a partial order are called reflexivity, antisymmetry, and
transitivity, respectively. Pairs of elements a,b G A such that neither a < b
nor b < a holds are said to be incomparable. One of the simplest examples
of a poset is the Boolean algebra Bn = (A, C), where A is all subsets of
{1,2,..., n} ordered by inclusion. In Bn^ the subsets {1} and {2} are incom-
parable. The set {0, 1, ... , n} with the normal ordering is an example of a
totally ordered set called the n-chain and denoted by Cn-
The particular partial order in which we are interested is the following.
Definition 2.2.2 Suppose A = (Ai, A 2 , . . . , A/) and pi = (jUi, /i 2 , • • • , Mm) are
partitions of n. Then A dominates pt^ written A > /i?
Ai + A2 + • • • "h A* ^ Ml d" M2 ’ * ■ “I" Mi
for alH > 1. If z > / (respectively, i > m), then we take A^ (respectively. Mi)
to be zero. ■
Intuitively, A is greater than pt in the dominance order if the Ferrers diagram
of A is short and fat but the one for m is long and skinny. For example, when
n = 6, then (3, 3) > (2, 2, 1, 1) since 3>2, 3 + 3>2 + 2, etc. However, (3, 3)
and (4, 1, 1) are incomparable since 3 < 4 but 3 4- 3 > 4 + 1.
Any partially ordered set can be visualized using a Hasse diagram.
Definition 2.2.3 If (A, <) is a poset and 5, c G A, then we say that b is
covered by c (or c covers 6 ), written 5 -4 c (or c >- 6 ), if 6 < c and there
is no d G A with b < d < c. The Hasse diagram of A consists of vertices
representing the elements of A with an arrow from vertex b up to vertex c if
b is covered by c. ■
2.2. DOMINANCE AND LEXICOGRAPHIC ORDERING
59
The Hasse diagram for the partitions of 6 ordered by dominance is given next.
( 6 )
t
(5.1)
t
(4.2)
\
(32)
\
/'
(4,12
(3,2,1)
Z'
\
(3,13)
\
(23)
(2M^)
t
( 2 , 1 ^)
t
( 1 ®)
The fundamental lemma concerning the dominance order is as follows.
Lemma 2.2.4 (Dominance Lemma for Partitions) Let and be ta-
bleaux of shape X and fi, respectively. If, for each index i, the elements of row
i of s^ are all in different columns in t^, then X> /a.
Proof. By hypothesis, we can sort the entries in each column of t^ so that
the elements of rows 1, 2, . . . , i of all occur in the first i rows of t^. Thus
Ai + A 2 H + Aj = number of elements in the first i rows of t^
> number of elements of s^ in the first i rows of t^
= /Xl + /i2 T • • • + ■
The second ordering on partitions is the one that would be given to them
in a dictionary.
Definition 2.2.5 Let A = (Ai, A 2 , . . . , A;) and pi = (//i, /i 2 , • • • , A^m) be par-
titions of n. Then A < // in lexicographic order if, for some index i,
Xj = fij for j < i and Xi < pii. m
This is a total ordering on partitions. For partitions of 6 we have
(16) < (2, 14) < (22, 12) < (23) < (3, C) < (3, 2, 1)
<(32)<(4, 12)<(4,2)<(5,1)<(6).
The lexicographic order is a refinement of the dominance order in the
sense of the following proposition.
60 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
Proposition 2.2.6 // h n with A > /i^ then \> fi.
Proof. If A 7 ^ /i, then find the first index i where they differ. Thus ~
/^j J2]=i ^3 > N (since A [> //). So A^ > /i^. ■
It turns out that we will want to list the in dual lexicographic order —
i.e., starting with the largest partition and working down. (By convention,
the conjugacy classes are listed in the character table in the usual dictionary
order so as to start with (1^), which is the class of the identity.) Note that
our first module will then be M^'^\ which is one-dimensional with the trivial
action. Thus we have an irreducible module to start with, as promised.
In most of our theorems, however, we will use the dominance ordering in
order to obtain the stronger conclusion that A > /i rather than just A > //.
2.3 Specht Modules
We now construct all the irreducible modules of <Sn- These are the so-called
Specht modules, S^.
Any tableau naturally determines certain isomorphic copies of Young sub-
groups in Sn>
Definition 2.3.1 Suppose that the tableau t has rows i^i, • • • , and
columns Ci , C 2 , . . . , . Then
Ht = X Sr^ X • • • X Sri
and
Ct = Sci X Sc2 X - ' ^Ck
are the row-stabilizer and column- stabilizer of t, respectively. ■
If we take
4
3
2
( 2 . 1 )
then
and
Rt — <5{1,2,4} X <^{3,5}
Ct = *^{3,4} X X <S{2}-
Note that our equivalence classes can be expressed as {t} = In addition,
these groups are associated with certain elements of C[<Sn]. In general, given
a subset H C Sn, we can form the group algebra sums
and
= E,r6if ®gn(7r)7T.
2.3. SPECHT MODULES
61
(Here, the group algebra will be acting on the permutation modules M^.
Thus the elements of C[<Sn] are not playing the roles of vectors and are not
set in boldface.) For a tableau the element Rf is already implicit in the
corresponding tabloid by the remark at the end of the previous paragraph.
However, we will also need to make use of
Kt ^ sgn(7r)7r.
7r€Ct
Note that if t has columns Ci, C 2 , . . . , then Kt factors as
^C2 • • • •
Finally, we can pass from t to an element of the module by the following
definition.
Definition 2.3.2 If t is a tableau, then the associated polytabloid is
e« = ■
To illustrate these concepts using the tableau t in (2.1), we compute
Kt = (e-(3,4))(€-(l,5)),
and so
_ 4 1 2 3 1 2 4 5 2 3 5 2
“ 3 5 4 5 3 1 4 1
The next lemma describes what happens to the various objects defined
previously in passing from ^ to 7rt.
Lemma 2.3.3 Let t be a tableau and tt be a permutation. Then
1. = 7ri?£7T“^
2. = 7^C(7r“^
3. = 7r/«t7r“\
4. = Tret.
of equivalent statements:
(j{'Kt} = {'Kt}
K~^(JK{t} = {t}
K~^(JK G Rt
(7 G kRiK~^ .
The proofs of parts 2 and 3 are similar to that of part 1.
Proof. 1. We have the following list
G G -^Trt ^ ^
62 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
4. We have
= 7T/^i{t} = Tret. ■
One can think of this lemma, for example part 1, as follows. If ^ has entries
then Tvt has entries nti^j. Thus an element of the row-stabilizer of irt may be
constructed by first applying 7 t“^ to obtain the tableau t, then permuting the
elements within each row of and finally reapplying tt to restore the correct
labels. Alternatively, we have just shown that if cr G Rnt^ then 7r“ Vtt G Rt
and the following diagram commutes:
a
TTt — > nti
7T~^ I t tt
t — > tl
7T“^(J7r
Finally, we are in a position to define the Specht modules.
Definition 2.3.4 For any partition A, the corresponding Specht module, S^,
is the submodule of M^ spanned by the polytabloids e^, where t is of shape
A. ■
Because of part 4 of Lemma 2.3.3, we have the following.
Proposition 2.3.5 The S^ are cyclic modules generated by any given poly-
tabloid. m
Let us look at some examples.
Example 2.3.6 Suppose A = (n). Then ei 2 ••• n = 1 2 • • n is the only
polytabloid and S^^^ carries the trivial representation. This is, of course, the
only possibility, since S^'^^ is a submodule of M^'^^ where Sn acts trivially
(Example 2.1.6). ■
Example 2.3.7 Let A = (1^) and fix
1
2
t= . . (2.2)
n
creSn
Thus
2.4. THE SUBMODULE THEOREM
63
and Ct is the signed sum of all n! permutations regarded as tabloids. Now for
any permutation tt, Lemma 2.3.3 yields
= Tret = ^ {sgna)-Ka{t}.
aeSn
Replacing Tra by r,
e^t = ^ (sgnTT" V)r{t} = (sgn7r~^) (sgnr)r{f} = (sgnTr)et
reSn reSn
because sgn7r“^ = sgnTr. Thus every polytabloid is a scalar multiple of e^,
where t is given by equation (2.2). So
5(1”) = C{et}
with the action Tre^ = (sgnTr)et. This is the sign representation. ■
Example 2.3.8 If A = (rz — 1, 1), then we can use the module isomorphism
of Example 2.1.8 to abuse notation and write (n — 1, l)-tabloids as
This tabloid has et = k — i, and the span of all such vectors is easily seen to
be
g{n 1 , 1 ) _ \ _j_ ^22 + • • * + Cji^ • C\ C2 Cji — 0 }.
So dim5^^~^’^^ = n — 1, and we can choose a basis for this module, e.g.,
6 = {2-l,3-l,...,n-l}.
Computing the action of tt G 5n on S, we see that the corresponding character
is one less than the number of fixedpoints of tt. Thus is the module
found at the end of Example 1.9.5. ■
The reader should verify that when n = 3, the preceding three examples
do give all the irreducible representations of 5s (which were found in Chapter
1 by other means). So at least in that case we have fulfilled our aim.
2.4 The Submodule Theorem
It is time to show that the 5^ constitute a full set of irreducible 5„-modules.
The crucial result will be the submodule theorem of James [Jam 76]. All the
results and proofs of this section, up to and including this theorem, are true
over an arbitrary field. The only change needed is to substitute a bilinear
form for the inner product of equation (2.3). The fact that this replaces
64 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
linearity by conjugate linearity in the second variable is not a problem, since
we never need to carry a nonreal scalar from one side to the other.
Recall that H~ = subset H C Sn- If H = {tt},
then we write 7t~ for H~ . We will also need the unique inner product on M^
for which
{^}) = (2.3)
Lemma 2.4.1 (Sign Lemma) Let H <Sn he a subgroup.
1. If 7T e H , then
7tH~ = H~7t = (sgnTr)^^-.
Otherwise put: H~ = H~ .
2. For any u, v G
{H-u,v) = {u,H-v).
3. If the transposition (6, c) G H, then we can factor
H- = k{e-{b,c)),
where k G C[<Sn].
4. If t is a tableau with 6, c in the same row of t and (6, c) G H, then
H~{t} = 0.
Proof. 1. This is just like the proof that net = (sgnTr)et in Example 2.3.7.
2. Using the fact that our form is «Sn-invariant,
(i/“u,v) = ^((sgn7r)7TU,v) = ^ (u, (sgn7r)7r“^v).
ttG// tzEH
Replacing tt by tt”^ and noting that this does not affect the sign, we see that
this last sum equals (u, ^f“v).
3. Consider the subgroup K = {e, (6, c)} of H. Then we can find a
transversal and write H = [^-kiK. But then H~ = (^ik~){e — (^, c)), as
desired.
4. By hypothesis, (6, c){t} = {t}. Thus
H~{t} = k{e - (6, c)){t} = k{{t} - {<}) = 0. ■
The sign lemma has a couple of useful corollaries.
Corollary 2.4.2 Let t = t^ be a X-tahleau and s = s^ be a p-tableau, where
A, // h n. If { 5 } / 0, then A > And if X = p, then = ±6^.
2.4. THE SUBMODULE THEOREM
65
Proof. Suppose b and c are two elements in the same row of s^. Then they
cannot be in the same column of for if so, then Kt = k{e — {b,c)) and
Kt{s} = 0 by parts 3 and 4 in the preceding lemma. Thus the dominance
lemma (Lemma 2.2.4) yields A > /.l
If A = /i, then we must have {s} = 7r{t} for some tt G Ct by the same
argument that established the dominance lemma. Using part 1 yields
= (sgn7r)Kt{t} == ±et. ■
Corollary 2.4.3 If u e M^ and sht = fi, then KtU is a multiple of et.
Proof. We can write u = where the Si are //-tableaux. By the
previous corollary, ■
We are now in a position to prove the submodule theorem.
Theorem 2.4.4 (Submodule Theorem [Jam 76]) Let U be a submodule
ofM^. Then
or U C
In particular, when the field is C, the are irreducible.
Proof. Consider u e U and a //-tableau t. By the preceding corollary, we
know that k^u = fet for some field element /. There are two cases, depending
on which multiples can arise.
Suppose that there exits a u and a t with / 7^ 0. Then since u is in the
submodule U, we have fst = KtU e U. Thus et ^U (since / is nonzero) and
CU (since 5^ is cyclic).
On the other hand, suppose we always have Ktxi = 0 . We claim that this
forces U C Consider any u G C/. Given an arbitrary //-tableau t, we
can apply part 2 of the sign lemma to obtain
(u, et) = (u, nt{t}) = (/^/u, {t}) = (0, {t}) = 0.
Since the et span 5^, we have u G 5^-^, as claimed. ■
It is only now that we will need our field to be the complexes (or any field
of characteristic 0).
Proposition 2.4.5 Suppose the field of scalars is C and 6 G Hom(S''^,M^)
is nonzero. Thus A > //, and if X = fi, then 9 is multiplication by a scalar.
Proof. Since 9^0, there is some basis vector et such that 9{et) 7^ 0. Because
(•, •) is an inner product with complex scalars, ® 5^-^. Thus we can
extend 9 to an element of Hom(M^, M^) by setting 9{S^-^) = 0. So
0 ^ 0{et) = 0{Kt{t}) = KtO{{t}) =
i
where the Si are //-tableaux. By Corollary 2.4.2 we have A > //.
66 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
In the case A Corollary 2.4.3 yields 0{et) = cst for some constant c.
So for any permutation tt,
0{e^t) = 0{7ret) = 7r0(et) = 7r{cet) = ce^t-
Thus 6 is multiplication by c. ■
We can finally verify all our claims about the Specht modules.
Theorem 2.4.6 The for X \~ n form a complete list of irreducible Sn~
modules over the complex field.
Proof. The S^ are irreducible by the submodule theorem and the fact that
S\ PI 5A_l _ Q (Q
Since we have the right number of modules for a full set, it suffices to show
that they are pairwise inequivalent. But if S^ = then there is a nonzero
homomorphism 9 G Hom(*S^, M^), since C M^. Thus \>ii (Proposition
2.4.5). Similarly, // > A, so A = /i. ■
Although the Specht modules are not necessarily irreducible over a field
of characteristic p, p prime, the submodule theorem says that the quotient
S^/{S^r\S^^) is. These are the objects that play the role of S^ in the theory
of p-modular representations of Sn> See James [Jam 78] for further details.
Corollary 2.4.7 The permutation modules decompose as
= 0
X> M
with the diagonal multiplicity = 1 .
Proof. This result follows from Proposition 2.4.5. If 5^ appears in with
nonzero coefficient, then A > //. If A = /i, then we can also apply Proposition
1.7.10 to obtain
= dimHom(S'^, M^) = 1. ■
The coefficients m\^ have a nice combinatorial interpretation, as we will see
in Section 2.11.
2.5 standard Tableaux and a Basis for
In general, the polytabloids that generate S^ are not independent. It would
be nice to have a subset forming a basis — e.g., for computing the matrices
and characters of the representation. There is a very natural set of tableaux
that can be used to index a basis.
Definition 2.5.1 A tableau t is standard if the rows and columns of t are
increasing sequences. In this case we also say that the corresponding tabloid
and polytabloid are standard. ■
2.5. STANDARD TABLEAUX AND A BASIS FOR S^
67
For example,
1 2 3
4 6
5
is standard, but
1 2 3
5 4
6
is not.
The next theorem is true over an arbitrary field.
Theorem 2.5.2 The set
{et : t is a standard X-tableau }
is a basis for .
We will spend the next two sections proving this theorem. First we will
establish that the et are independent. As before, we will need a partial order,
this time on tabloids.
It is convenient at this point to consider ordered partitions.
Definition 2.5.3 A composition of n is an ordered sequence of nonnegative
integers
A = (Ai, A2, . . . , Ai)
such that ^ • Xi = n. The integers are called the parts of the composition.
Note that there is no weakly decreasing condition on the parts in a compo-
sition. Thus (1,3,2) and (3,2,1) are both compositions of 6, but only the
latter is a partition. The definitions of a Ferrers diagram and tableau are
extended to compositions in the obvious way. (However, there are no stan-
dard A-tableaux if A is not a partition, since places to the right of or below
the shape of A are considered to be filled with an infinity symbol. Thus
columns do not increase for a general composition.) The dominance order
on compositions is defined exactly as in Definition 2.2.2, only now Ai, . . . , A^
and pi, . . . are just the first i parts of their respective compositions, not
necessarily the i largest.
Now suppose that {t} is a tabloid with sht = A h n. For each index i,
1 < 2 < n, let
{T} = the tabloid formed by all elements < 2 in {t}
and
A* = the composition which is the shape of {t^}
68 CHAPTER 2, REPRESENTATIONS OF THE SYMMETRIC GROUP
As an example, consider
Then
A' =(0,1), A2
{t}
2
4
1
CO
1 ’
2
1 3 ’
(1,1), A3 =(1,2),
2 4
1 3 ’
A^ =(2,2).
The dominance order on tabloids is determined by the dominance ordering
on the corresponding compositions.
Definition 2.5.4 Let {s} and {t} be two tabloids with composition se-
quences A® and /x®, respectively. Then {s} dominates {t}, written {«} > {t},
if A® > for all i. m
The Hassc diagram for this ordering of the (2, 2)-tabloids is as follows;
2 3
1 4
\
1
2
3
4
t
1
3
2
4
2
4
1
3
t
3
4
1
2
\
1 4
2 3
Just as for partitions, there is a dominance lemma for tabloids.
Lemma 2.5.5 (Dominance Lemma for Tabloids) If k < I and k ap-
pears in a lower row than I in {t}, then
{t} < {k,l){t}.
Proof. Suppose that {t} and {k, l){t} have composition sequences and /i^
respectively. Then lov i < k ov i > I we have
Now consider the case where k < i < 1. If r and q are the rows of {^} in
which k and I appear, respectively, then
A^ = with the ^th part decreased by 1
and the rth part increased by 1.
2.5. STANDARD TABLEAUX AND A BASIS FOR S^
69
Since g < r by assumption, A* <1 ■
If V = T,iCi{ti} G M^, then we say that {U} appears in \ Ci ^ 0.
The dominance lemma puts a restriction on which tableaux can appear in a
standard polytabloid.
Corollary 2.5.6 If t is standard and {s} appears in St, then {t} > {s}.
Proof. Let s = ttL where tt G so that {s} appears in e^. We induct on
the number of column inversions in 5, i.e., the number of pairs A; < / in the
same column of s such that /c is in a lower row than I, Given any such pair,
{s} < {kj){s}
by Lemma 2.5.5. But (/c, l){s} has fewer inversions than {s}, so, by induction,
(fc,/){s} < {t} and the result follows. ■
The previous corollary says that {t} is the maximum tabloid in e^, by
which we mean the following.
Definition 2.5.7 Let (A, <) be a poset. Then an element b e A is the
maximum \ih>c for all c G A. An element 6 is a maximal element if there is
no c G A with c> b. Minimum and minimal elements are defined analogously.
Thus a maximum element is maximal, but the converse is not necessarily
true. It is important to keep this distinction in mind in the next result.
Lemma 2.5.8 Let vi, V 2 , . . . , Vm be elements of M^. Suppose, for each Vi,
we can choose a tabloid {ti} appearing in such that
1. {ti} is maximum in \i, and
2. the {ti} are all distinct.
Then vi , V 2 , . . . , are independent.
Proof. Choose the labels such that {ti} is maximal among the {U}. Thus
conditions 1 and 2 ensure that {ti} appears only in Vi. (If {ti} occurs in
Vi, i > 1, then contradicting the choice of {ti}.) It follows that
in any linear combination
CiVi 4- C2V2 H + CrnVm = 0
we must have ci = 0 because there is no other way to cancel {ti}. By
induction on m, the rest of the coefficients must also be zero. ■
The reader should note two things about this lemma. First of all, it is
not sufficient only to have the {U} maximal in v^; it is easy to construct
counterexamples. Also, we have used no special properties of > in the proof,
so the result remains true for any partial order on tabloids.
We now have all the ingredients to prove independence of the standard
basis.
70 CHAPTER 2, REPRESENTATIONS OF THE SYMMETRIC GROUP
Proposition 2.5.9 The set
{e^ : t is a standard X- tableau}
is independent.
Proof. By Corollary 2.5.6, {t} is maximum in e^, and by hypothesis they are
all distinct. Thus Lemma 2.5.8 applies. ■
2.6 Garnir Elements
To show that the standard poly tabloids of shape p span 5^, we use a pro-
cedure known as a straightening algorithm. The basic idea is this. Pick an
arbitrary tableau t. We must show that et is a linear combination of standard
polytabloids. We may as well assume that the columns of t are increasing,
since if not, there is cr € Ct such that s = at has increasing columns. So
€s = aet = {sgn a)et
by Lemmas 2.3.3 (part 4) and 2.4.1 (part 1). Thus et is is a linear combination
of poly tabloids whenever €« is.
Now suppose we can find permutations tt such that
1. in each tableau nt, a certain row descent of t (pair of adjacent, out-of-
order elements in a row) has been eliminated, and
2. the group algebra element g = e-\- Yl^{^gn'K)'K satisfies get = 0.
Then
^ ^ ♦
7T
So we have expressed et in terms of polytabloids that are closer to being
standard, and induction applies to obtain et as a linear combination of poly-
tabloids.
The elements of the group algebra that accomplish this task are the Garnir
elements.
Definition 2.6.1 Let A and B be two disjoint sets of positive integers and
choose permutations tt such that
Saub = y 7 t(«Sa X Sb)’
TT
Then a corresponding Garnir element is
9a,b = ^(sgn7r)7r. ■
7T
2.6. GARNIR ELEMENTS
71
Although qa,b depends on the transversal and not just on A and we will
standardize the choice of the tt’s in a moment. Perhaps the simplest way to
obtain a transversal is as follows. The group Saub acts on all ordered pairs
{A', B') such that \A'\ -- \ A\, \B'\ = \B\, and A' (jj jB' — A (+J 5 in the obvious
manner. If for each possible (A', B') we take tt G Saub such that
7r{A,B) = {A\B'),
then the collection of such permutations forms a transversal. For example,
suppose A = {5, 6} and B = (2, 4}. Then the corresponding pairs of sets (set
brackets and commas having been eliminated for readability) and possible
permutations are
(A', B') : (56, 24) , (46, 25) , (26, 45) , (45, 26) , (25, 46) , (24, 56).
9a^b= e - (4,5) - (2,5) - (4,6) - (2,6) + (2,5)(4,6).
It should be emphasized that for any given pair (A', 5'), there are many
different choices for the permutation tt sending (A, B) to that pair.
The Garnir element associated with a tableau t is used to eliminate a
descent Uj >
Definition 2.6.2 Let t be a tableau and let A and B be subsets of the jth
and {j T l)st columns of t, respectively. The Garnir element associated with
t (and A, B) is Qa.b = X^7r(sgn7r)7r, where the tt have been chosen so that
the elements of A U 5 are increasing down the columns of ttL ■
In practice, we always take A (respectively, B) to be all elements below Lj
(respectively, above as in the diagram
If we use the tableau
t =
1 2 3
5 4
6
with the descent 5 > 4, then the sets A and B are the same as in the previous
example. Each (A', B') has a corresponding t' that determines a permutation
in 9a,b'
123 123 143 123 143 153
t':54, 45, 25, 46, 26, 26,
6 6 6 5 5 4
9a,b^ e - (4,5) + (2,4,5) + (4,6,5) - (2, 4, 6, 5) + (2,5)(4,6).
72 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
The reader can verify that gA.B^t — 0? so that
— ^t2 “ ^t3 ~ + ^^5 ~ 5
where ^ 2 , • • • 5^6 are the second through sixth tableaux in the preceding list.
Note that none of these arrays have the descent found in the second row of t.
Proposition 2.6.3 Let t, A, and B, he as in the definition of a Garnir
element. If \AU B\ is greater than the number of elements in column j of t,
then gA,B^t = 0-
Proof. First, we claim that
= 0. (2.4)
Consider any a E Ct- By the hypothesis, there must he a,b e AUB such that
a and b are in the same row of at. But then (a, b) e Saub and ^aub{^^} ~ ®
by part 4 of the sign lemma (Lemma 2.4.1). Since this is true of every a
appearing in the claim follows.
Now Saub = l±) 7 r^(‘ 5 A x Sb), so S^^^ = gA,B{^A x Sb)~ • Substituting
this into equation (2.4) yields
gA,B{SA X SB)~^t = 0, (2.5)
and we need worry only about the contribution of {Sa x Sb)~- But we have
Sa X Sb Q Cf. So if a e Sa X Sb, then, by part 1 of the sign lemma.
a et = a C^. {t} = C^ {t} = Ct.
Thus {Sa X Sb) = \Sa x SB\et, and dividing equation (2.5) by this car-
dinality yields the proposition. ■
The reader may have noticed that when we eliminated the descent in row
2 of the preceding example, we introduced descents in some other places —
e.g., in row 1 of t^. Thus we need some measure of standardness that makes
/: 2 , . . . , ^6 closer to being standard than t. This is supplied by yet another par-
tial order. Given consider its column equivalence class^ or column tabloid.,
[t] = CtL
i.e., the set of all tableaux obtained by rearranging elements within columns
of t. We use vertical lines to denote the column tabloid, as in
1
3
2
1 2 3 2 1
3 ’ 1 /■
Replacing “row” by “column” in the definition of dominance for tabloids, we
obtain a definition of column dominance for which we use the same symbol
as for rows (the difference in the types of brackets used for the classes makes
the necessary distinction).
Our proof that the standard polytabloids span 5^ follows Peel [Pee 75].
2.6. GARNIR ELEMENTS
73
Theorem 2.6.4 The set
{st : t is a standard X-tableau} (2.6)
spans S^.
First note that if Ct is in the span of the set (2.6), then so is Cs for any
5 G [t] , by the remarks at the beginning of this section. Thus we may always
take t to have increasing columns.
The poset of column tabloids has a maximum element [^o], where to is
obtained by numbering the cells of each column consecutively from top to
bottom, starting with the leftmost column and working right. Since to is
standard, we are done for this equivalence class.
Now pick any tableau t. By induction, we may assume that every tableau
s with [s] t> [t] is in the span of (2.6). If t is standard, then we are done. If
not, then there must be a descent in some row i (since columns increase). Let
the columns involved be the jth and (j + l)st with entries ai < U2 < • • • < Up
and 6i < 62 < • • • < respectively. Thus we have the following situation in
t:
ai bi
A
tt2 62
A
A
Uj ^ bi
A
A bq
Up
Take A = {ai, . . . , Up} and B = {61 , . . . , 6 ^} with associated Garnir element
9a,b = Proposition 2.6.3 we have = 0, so that
et = - ^(sgn n)e^f (2.7)
But bi <•'•< bi < ai <• - < Qp implies that [irt] > [t] for tt ^ e by the
column analogue of the dominance lemma for tabloids (Lemma 2.5.5). Thus
all terms on the right side of (2.7), and hence et itself, are in the span of the
standard polytabloids. ■
To summarize our results, let
= the number of standard A-tableaux.
Then the following is true over any base field.
74 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
Theorem 2.6.5 For any partition X:
1. {et : t is a standard X- tableau} is a basis for S^,
2. dim5'^ = f^, and
3 . =
Proof. The first two parts are immediate. The third follows from the fact
(Proposition 1.10.1) that for any group G,
^(dimF)2 = |G|,
where the sum is over all irreducible G-modules. ■
2.7 Young’s Natural Representation
The matrices for the module in the standard basis form what is known as
Young ’s natural representation. In this section we illustrate how these arrays
are obtained.
Since Sn is generated by the transpositions (A:,A:-|-1) (or k = l,...,n — 1,
one need compute only the matrices corresponding to these group elements.
If t is a standard tableau, we get the tth column of the matrix for tt G by
expressing Tret = as a sum of standard polytabloids. When tt = (/c, /c-h 1),
there are three cases.
1. If k and /c + 1 are in the same column of t, then {k, k 1) e Ct and
{k,k+ l)et = -et.
2. If k and A; + 1 are in the same row of then (/c, k + l)t has a descent in
that row. Applying the appropriate Garnir element, we obtain
(/c, k 4- l)et = et zb other polytabloids e^/ such that [t'] D> [t].
3. If k and A: + 1 are not in the same row or column of i, then the tableau
t' = (k,k F 1)^ is standard and
(A:, A; 4- l)et = et>.
The et in the sum for case 2 comes from the term of equation (2.7) corre-
sponding to 7T = (fc. A; 4- 1). Although we do not have an explicit expression
for the rest of the terms, repeated application of 1 through 3 will compute
them (by reverse induction on the column dominance ordering). This is the
straightening algorithm mentioned at the beginning of Section 2.6.
2.7. YOUNG^S NATURAL REPRESENTATION
75
By way of example, let us compute the matrices for the representation of
S 3 corresponding to A == (2, 1 ). The two standard tableaux of shape A are
1 3
2
and t 2
1 2
3
Note that we have chosen our indices so that [ti] 1 > [^ 2 ]- This makes the
computations in case 2 easier.
For the transposition (1,2) we have
n o^ _ 2 3 1 3 _
( 1 , ^ 2
as predicted by case 1 . Since
(l,2)i2= 3^
has a descent in row 1, we must find a Garnir element. Taking A = {2,3}
and B = {1} gives tableaux
2 1 n 12 ^ ,13,
^ =(l, 2 )t 2 , 3 =^ 2 , and ^ ==
with element
= (1,2) + (1,3,2).
Thus
6(1, 2)^2 — ^t 2 + ^ti — O5
or
( 1 , 2 ) 6^2 = et^ - et,
(This result can be obtained more easily by merely noting that
(l, 2 )e,,
3 1
2
and expressing this as a linear combination of the e^..) Putting everything
together, we obtain the matrix
x(d,2))=(' ■; V
The transposition (2,3) is even easier to deal with, since case 3 always
applies, yielding
and
Thus
(2, S)ti = t 2
( 2 , 3)^2 =h.
X((2,3))= ( 5 I
We have already seen these two matrices in Example 1.9.5. Since the adjacent
transpositions generate Sn, all other matrices must agree as well.
76 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
2.8 The Branching Rule
It is natural to ask what happens when we restrict or induce an irreducible
representation of Sn to Sn-i or Sn-^i , respectively. The branching theorem
gives a simple answer to that question.
Intuitively, these two operations correspond to either removing or adding
a node to the Ferrers diagram for A.
Definition 2.8.1 If A is a diagram, then an inner corner 0 / A is a node
(i, j) G A whose removal leaves the Ferrers diagram of a partition. Any
partition obtained by such a removal is denoted by A“. An outer corner
0 / A is a node (z, j) 0 A whose addition produces the Ferrers diagram of a
partition. Any partition obtained by such an addition is denoted by A"^. ■
Note that the inner corners of A are exactly those nodes at the end of a
row and column of A. For example, if A = (5, 4, 4, 2), then the inner corners
are enlarged and the outer corners marked with open circles in the following
diagram:
• • • • # o
• • • • o
A == • • • •
• • o
o
So, after removal, we could have
A- :
• • • •
• • • •
• • • • ’
• •
• • • •
• • •
• •
• • • •
• • • •
whereas after addition, the possibilities are
A+:
• • • •
• • • •
• • • •
• • • •
• •
• • • •
• • • •
• • •
• • •
• • •
These are exactly the partitions that occur in restriction and induction. In
particular,
5(4, 4, 4, 2) ^ 5(5, 4, 3, 2) ^ 5(5, 4, 4,1)
and
^(5,4,4,2)|5 i 6^ ^(6, 4, 4, 2) 0 ^(5, 5, 4, 2) 0 ^(5, 4, 4, 3) 0 ^(5,4,4,2,1) ^
Before proving the branching rule itself, we need a result about dimen-
sions.
2.8. THE BRANCHING RULE
77
Lemma 2.8.2 We have
X-
Proof. Every standard tableau of shape A h n consists of n in some inner
corner together with a standard tableau of shape A“ h n — 1. The result
follows. ■
Theorem 2.8.3 (Branching Rule) If X\- n, then
1. ’ and
X-
2 .
A+
Proof. [Pee 75] 1. Let the inner corners of A appear in rows ri < V 2 <
• • • < For each i, let A^ denote the partition A" obtained by removing
the corner cell in row In addition, if n is at the end of row ri of tableau t
(respectively, in row Vi of tabloid {t}), then (respectively, {^*}) will be the
array obtained by removing the n.
Now given any group G with module V and submodule VF, it is easy to
see that
V^we{v/W),
where VjW is the quotient space. (See Exercise 7 in Chapter 1.) Thus it
suffices to find a chain of subspaces
{0} = c C C • • • C
such that = 5^’ as «S„_i-modules for 1 < z < A:. Let be
the vector space spanned by the standard polytabloids e^, where n appears
in t at the end of one of rows ri through ri. We show that the are our
desired modules as follows.
Define maps 6i : by linearly extending
//I ^ ^
^ (0 otherwise.
The reader can quickly verify that 6i is an <Sn-i-homomorphism. Further-
more, for standard t we have
^ ^ f if n is in row ri of t,
I 0 if n is in row rj of t, where j < i.
This is because any tabloid appearing in e^, ^ standard, has n in the same
row or higher than in t.
Since the standard polytabloids form a basis for the corresponding Specht
module, the two parts of (2.8) show that
(2.9)
78 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
and
Cker6»i.
( 2 . 10 )
From equation (2.10), we can construct the chain
{ 0 } = C 1 /( 1 ) n ker 6>1 C l/d) C 1 /( 2 ) n ker6i2 C y( 2 ) C • • • C !/('') = S^.
( 2 . 11 )
But from equation (2.9)
dim
y(0
1/(0 n ker
By the preceding lemma, the dimensions of these quotients add up to dim S^.
Since this leaves no space to insert extra modules, the chain (2.11) must have
equality for the first, third, etc. containments. Furthermore,
y(0
1 / 0 - 1 ) ~
1 / 0 )
F(0 n ker^i
as desired.
2. We will show that this part follows from the first by Frobenius reci-
procity (Theorem 1.12.6). In fact, parts 1 and 2 can be shown to be equivalent
by the same method.
Let be the character of 5^. If then by taking
characters, The multiplicities are given by
/ih-n+l
rrif, =
= <x^.x^4-5„)
= )
r 1 i^ X = ,
\ 0 otherwise
(Corollary 1.9.4, part 2)
(Frobenius reciprocity)
(branching rule, part 1)
(Corollary 1.9.4, part 4)
( 1 if /i = A"*",
\ 0 otherwise.
(definition of /i“ and A“t)
This finishes the proof. ■
2.9 The Decomposition of
We would like to know the multiplicity m\^ of the Specht module S^ in AH.
In fact, we can give a nice combinatorial description of these numbers in terms
of tableaux that allow repetition of entries.
Definition 2.9.1 A generalized Young tableau of shape A, is an array T ob-
tained by replacing the nodes of A with positive integers, repetitions allowed.
The type or content of T is the composition /i = (/ii,// 2 , • • • ^Mm), where
equals the number of i’s in T. Let
TA/i = {T : T has shape A and content /i}. ■
2.9. THE DECOMPOSITION OF
79
Note that capital letters will be used for generalized tableaux. One such
array is
^^414
1 3
of shape (3, 2) and content (2, 0, 1, 2).
We will show that C[7 a^] is really a new copy of M^. For the rest of this
section and the next:
fix a tableau t of shape X and content (1’^).
In all our examples we will use A = (3, 2) and
If T is any A-tableau, then let T{i) denote the element of T in the same
position as the i in the fixed tableau t. With t and T as before,
T(l) T{2) T(3)
T(4) T(5)
( 2 . 12 )
so that T(l) = T(3) = 4, T(2) = T(4) = 1, and T(5) = 3.
Now given any tabloid {s} of shape /z, produce a tableau T G T\^ by
letting
T(i) = the number of the row in which i appears in {s}.
For example, suppose /i = (2, 2, 1) and
2 3
W = J_l_.
4
Then the 2 and 3 are in row one of {s}, so the 2 and 3 of t get replaced by
ones, etc., to obtain
^^211
3 2'
Note that the shape of {s} becomes the content of T, as desired. Also, it is
clear that the map {«} -> T is a bijection between bases for and CfTA/x],
so they are isomorphic as vector spaces.
We now need to define an action of Sn on generalized tableaux so that
= C[T\^] as modules. If tt G <Sn and T G then we let ttT be the
tableau such that
(7rT)(i)
To illustrate with our canonical tableau (2.12):
(1.2, 4)T
T(4) T(l) T(3)
T(2) T(5)
80 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
In particular,
(1,2,4) ^ 2
1 _ 3 2
1 2
1
Note that although tt e Sn acts on the elements of {s} (replacing i by 7ri),
it acts on the places in T (moving T(i) to the position of T{7ri)). To see
why this is the correct definition for making 6 into a module isomorphism,
consider ^({s}) = T and tt G Sn- Thus we want 0(7 t{s}) = ttT. But this
forces us to set
(7tT)( 2) = row number of i in 7t{s}
= row number of in {s}
= T{7T-H).
We have proved, by fiat, the following.
Proposition 2.9.2 For any given partition X, the modules and C[7a/x]
are isomorphic, m
Recall (Proposition 1.7.10) that the multiplicity of in is given by
dimHom(5'^,M^). We will first construct certain homomorphisms from
to using our description of the latter in terms of generalized tableaux and
then restrict to 5^. The row and column equivalence classes of a generalized
tableau T, denoted by {T} and [T], respectively, are defined in the obvious
way. Let {t} G be the tabloid associated with our fixed tableau.
Definition 2.9.3 For each T G the homomorphism corresponding to T
is the map 6t G Hom(M^, M^) given by
se{T}
and extension using the cyclicity of M^. Note that 6t is actually a homomor-
phism into C7 a^, but that should cause no problems in view of the previous
proposition. ■
Extension by cyclicity means that, since every element of is of the
form g{t} for some g G C[5n], we must have Origit}) = 9Y^se{T} ^ (fh^f
6t respects the group action and linearity). For example, if
2 1
3 2
then
6r{t} =
2 11
3 2
+
12 1
3 2
112
3 2
+
2 11
2 3
+
12 1 112
2 3 ^23
and
0T(l,2,4){t}
3 2 1
1 2
311 312221211
22 ^12 ^13 ^23
+
2 12
1 3
2.9. THE DECOMPOSITION OF
81
Now we obtain elements of Hom(5^, M^) by letting
9t = the restriction of 6t to S^.
If t is our fixed tableau, then
se{T}
This last expression could turn out to be zero (thus forcing 0t to be the zero
map by the cyclicity S^) because of the following.
Proposition 2.9.4 If t is the fixed X-tableau and T G T\^, then ntT = 0 if
and only ifT has two equal elements in some column.
Proof. If ntT = 0, then
T + ^ (sgn7r)7rT = 0.
neCt
TT^e.
So we must have T = ttT for some n E Ct with sgnTr = — 1. But then the
elements corresponding to any nontrivial cycle of tt are all equal and in the
same column.
Now suppose that T{i) = T{j) are in the same column of T. Then
{e-{iJ))T = 0.
But e — {i,j) is a factor of nt by part 3 of the sign lemma (Lemma 2.4.1), so
KtT = 0 . ■
In light of the previous proposition, we can eliminate possibly trivial 6t
from consideration by concentrating on the analogue of standard tableaux for
arrays with repetitions.
Definition 2.9.5 A generalized tableau is semistandard if its rows weakly
increase and its columns strictly increase. We let 7^^ denote the set of semi-
standard A-tableaux of type /x. ■
The tableau
1 2
3
is semistandard, whereas
2 1
3 2
1
is not. The homomorphisms corresponding to semistandard tableaux are the
ones we have been looking for. Specifically, we will show that they form a
basis for Hom(5^, M^).
82 CHAPTER 2. REPRESENTATIONS OE THE SYMMETRIC GROUP
2.10 The Semistandard Basis for Hom(6'^, M^)
This section is devoted to proving the following theorem.
Theorem 2.10.1 The set
{St : TeT^^}
is a basis for Hom(5^, M^).
In many ways the proof will parallel the one given in Sections 2.5 and 2.6
to show that the standard poly tabloids form a basis for S^.
As usual, we need appropriate partial orders. The dominance and column
dominance orderings for generalized tableaux are defined in exactly the same
way as for tableaux without repetitions (see Definition 2.5.4). For example,
if
[5]=|2|1|1| and [T] =
then [S] corresponds to the sequence of compositions
A^ = (0,1,1), A^ = (1,2,1), A^ = (2,2,1),
whereas [T] has
m' = (1,1,0), = (2,1,1), = (2,2,1).
Since A* < /i* for all i, [S] < [T]. The dominance lemma for tabloids (Lemma
2.5.5) and its corollary (Corollary 2.5.6) have analogues in this setting. Their
proofs, being similar, are omitted.
Lemma 2.10.2 (Dominance Lemma for Generalized Tableaux) Let k
be in a column to the left of I in T with k <1. Then
[T] t> [5],
where S is the tableau obtained by interchanging k and I in T. m
Corollary 2.10.3 If T is semistandard and S G {T} is different from T,
then
[T] t> [5], .
Thus, if T is semistandard, then [T] is the largest equivalence class to appear
in Orit}.
Before proving independence of the Ot, we must cite some general facts
about vector spaces. Let F be a vector space and pick out a fixed basis
B = {bi, &25 • • • 5 bn}- If V G V', then we say that bi appears m v if v = Yli ^ibi
with Ci ^ 0. Suppose that V is endowed with an equivalence relation whose
equivalence classes are denoted by [v], and with a partial order on these
classes. We can generalize Lemma 2.5.8 as follows.
1 1 2 I
2 3
2.10. THE SEMISTANDARD BASIS FOR H0M(5^,
83
Lemma 2.10.4 Let V and B he as before and consider a set of vectors
vi, V 2 , . . . , v^, G V. Suppose that, for all i, there exists bi e B appearing
in \i such that
1- [^i] ^ [b] for every b ^ b{ appearing in Wi, and
2. the [bi] are all distinct.
Then the v* are linearly independent, m
We also need a simple lemma about independence of linear transforma-
tions.
Lemma 2.10.5 Let V and W he vector spaces and let ^ 2 ? • • • ? linear
maps from V to W. If there exists a v such that ^i(v), 02(v), . . . , 0m(v)
are independent in W, then the 6i are independent as linear transformations.
Proof. Suppose not. Then there are constants c^, not all zero, such that
CiOi is the zero map. But then Y2- CiOiiy) = 0 for all v G V, a contradic-
tion to the hypothesis of the lemma. ■
Proposition 2.10.6 The set
{Ot :
is independent.
Proof. Let Ti, T 2 , . . . , Tjn be the elements of 7 a^. By the previous lemma, it
suffices to show that OtiGi, 6 x 2 ^ 1 ^ • • • are independent, where t is our
fixed tableau. For all i we have
Now Ti is semistandard, so [T^] t> [5] for any other summand 5 in 0Ti{t}
(Corollary 2.10.3). The same is true for summands of since the
permutations in Kt do not change the column equivalence class. Also, the [Ti]
are all distinct, since no equivalence class has more than one semistandard
tableau. Hence the satisfy the hypotheses of Lemma 2.10.4,
making them independent. ■
To prove that the Or span, we need a lemma.
Lemma 2.10.7 Consider 6 G Hom(*S'^, M^). Write
Get = ^ ctT,
T
where t is the fixed tableau of shape A.
1. If n e Ct and T\ = 7tT2, then ct^ = (sgn7r)cT2-
84 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
2. Every T\ with a repetition in some column has Ct^ =0.
3. If 0 is not the zero map, then there exists a semistandard T2 having
ct 2 7^ 0-
Proof. 1. Since tt ^ Ct, we have
n{eet) = 0{iTKt{t}) = 0((sgn7r)Kt{f}) = (sgn7r)(6>et).
Therefore, Get = implies
w^ctT == niGet) = (sgn7r)(0et) = (sgnTr) ^ctT.
T T
Comparing coefficients of 7 tT 2 on the left and T\ on the right yields =
(sgnTr)cTi, which is equivalent to part 1.
2. By hypothesis, there exists {i,j) G Ct with {i,j)T\ = T\. But then
cti = — cti by what we just proved, forcing this coefficient to be zero.
3. Since 0 ^ 0, we can pick T2 with ct2 7^ 0 such that [T2] is maximal.
We claim that T2 can be taken to be semistandard. By parts 1 and 2, we can
choose T2 so that its columns strictly increase.
Suppose, toward a contradiction, that we have a descent in row i. Thus
T2 has a pair of columns that look like
hi
A
b2
A
A
ai > hi
A
A
Op
Choose A and B as usual, and let qa^b = associated
Garnir element. We have
9 a,b(^ctT) = gA,B{0^t) = 0{gA,B^t) = ^(0) = 0-
T
Now T2 appears in gA,BT2 with coefficient 1 (since the permutations in pa,b
move distinct elements of T2). So to cancel T2 in the previous equation, there
must be a T ^ T2 with ttT = T2 for some tt in gA.B- Thus T is just T2 with
some of the a^’s and b^s exchanged. But then [T] D> [T2] by the dominance
lemma for generalized tableaux (Lemma 2.10.2). This contradicts our choice
of T2. ■
We are now in a position to prove that the 9 t generate Hom(5^, M^).
2.11. KOSTKA NUMBERS AND YOUNG’S RULE
85
Proposition 2.10.8 The set
{6 t : Tg73°^}
spans Hom(5^, M^).
Proof. Pick any 6 G Hom(5^,M^) and write
eet = Y,<^rT.
T
(2.13)
Consider
Le = {S e 7x^ : [5] < [T] for some T appearing in Set}.
In poset terminology, Lq corresponds to the lower order ideal generated by
the T in Oet. We prove this proposition by induction on \Lq\.
If I Z/( 9 1 =0, then 6 is the zero map by part 3 of the previous lemma. Such
a ^ is surely generated by our set!
If \Le\ > 0, then in equation (2.13) we can find a semistandard T 2 with
ct 2 ^ 0. Furthermore, it follows from the proof of part 3 in Lemma 2.10.7
that we can choose [T 2 ] maximal among those tableaux that appear in the
sum. Now consider
62 = 0 — CT 2 ^T2 •
We claim that is a subset of Lq with T2 removed. First of all, every S
appearing in 0T2^t satisfies [5']<[T2] (see the comment after Corollary 2.10.3),
so Lq 2 S Lq. Furthermore, by part 1 of Lemma 2.10.7, every S with [S] = [T2]
appears with the same coefficient in Oet and Thus T2 ^ 1/^2 , since
[T 2 ] is maximal. By induction, 62 is in the span of the 6 t and thus 6 is as
well.
This completes the proof of the proposition and of Theorem 2.10.1. ■
2.11 Kostka Numbers and Young’s Rule
The Kostka numbers count semistandard tableaux.
Definition 2.11.1 The Kostka numbers are
Kxf. = | 7 )°,|. ■
As an immediate corollary of the semistandard basis theorem (Theorem
2.10.1), we have Young’s rule.
Theorem 2.11.2 (Young’s Rule) The multiplicity of in is equal
to the number of semistandard tableaux of shape X and content p, i.e.,
A
86 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
Note that by Corollary 2.4.7, we can restrict this direct sum to A > /u. Let us
look at some examples.
Example 2.11.3 Suppose // = (2,2,1). Then the possible A > ju and the
associated A- tableaux of type fi are as follows:
' = (2,2,1)
A2= (3,1,1)
A3= (3,2)
A^ = (4, 1) A^
= (5)
• •
• • •
• « •
• • • •
= 9 •
= •
= •
=
9
•
Till
1 1 2
1 1 2
112 2
112 2 3
2 2
2
2 3
3
3
3
1 1 3
112 3
2 2
2
Thus
j^(2,2,i) ^ 5(2,2,!) 0 5(3,3,!) 0 0 25 ^^’^^ 0
Example 2.11.4 For any //, = 1. This is because the only /i-tableau
of content fi is the one with all the I’s in row 1, all the 2’s in row 2, etc. Of
course, we saw this result in Corollary 2.4.7.
Example 2.11.5 For any fi, = L Obviously there is only one way to
arrange a collection of numbers in weakly increasing order. It is also easy to
see from a representation-theoretic viewpoint that must contain exactly
one copy of (see Exercise 5b).
Example 2.11.6 For any A, JTa(!^) = (fhe number of standard tableaux
of shape A). This says that
A
But is just the regular representation (Example 2.1.7) and = dim 5^
(Theorem 2.5.2). Thus this is merely the special case of Proposition 1.10.1,
part 1, where G = Sn>
2.12 Exercises
1 . Let be the character of M^. Find (with proof) a formula for the
value of on the conjugacy class Kx-
2. Verify the details in Theorem 2.1.12.
2.12. EXERCISES
87
3. Let A = (Ai, A 2 , . . . , A/) and // = , fJ^m) be partitions. Char-
acterize the fact that A is covered by /i if the ordering used is
(a) lexicographic,
(b) dominance.
4. Consider where each tabloid is identified with the element in
its second row. Prove the following facts about this module and its
character.
(a) We have
g{n 1,1) _ _l_ q^2 -!-••• + c^n : Ci + C 2 H- • • • + = 0}.
(b) For any tt G 5^,
x(’^~bi)( 7 r) = (number of fixedpoints of tt) — 1.
5. Let the group G act on the set S. We say that G is transitive if, given
any s, ^ G 5, there is a ^ G G with gs = t. The group is doubly transitive
if, given any s^t^u^v G S with s ^ u and t ^ v^ there is a ^ G G with
gs — t and gu — v. Show the following.
(a) The orbits of G’s action partition S.
(b) The multiplicity of the trivial representation m. V = CS is the
number of orbits. Thus if G is transitive, then the trivial represen-
tation occurs exactly once. What does this say about the module
Mn
(c) If G is doubly transitive and V has character x? then x — 1 is an
irreducible character of G. Hint: Fix s G 5 and use Frobenius
reciprocity on the stabilizer Gs < G.
(d) Use part (c) to conclude that in Sn the function
/(tt) = (number of fixedpoints of tt) - 1
is an irreducible character.
6. Show that every irreducible character of Sn is an integer- valued func-
tion.
7. Define a lexicographic order on tabloids as follows. Associate with any
{t} the composition A = (Ai, A 2 , • . . , A^), where A^ is the number of the
row containing n — i 1. If {s} and {t} have associated compositions
A and //, respectively, then {s} < {t} in lexicographic order (also called
last letter order) if A < /i.
(a) Show that {s} < {t} implies {s} < {t}.
(b) Repeat Exercise 3 with tabloids in place of partitions.
88 CHAPTER 2. REPRESENTATIONS OF THE SYMMETRIC GROUP
8. Verify that the permutations tt chosen after Definition 2.6.1 do indeed
form a transversal for Sa x Sb in Saub-
9. Verify the statements made in case 2 for the computation of Young’s
natural representation (page 74).
10. In Sn consider the transpositions Tk = {k^k 1) for A: = 1, . . . , n — 1.
(a) Prove that the generate Sn subject to the Coxeter relations
rt = e,
1 < A; < n — 1,
) 1 ^ /c ^ 71 2,
Tj^n = TiTk, 1 < fc, / < n — 1 and |A; — ^| > 2.
(b) Show that if Gn is a group generated by for A: = 1, . . . , n — 1
subject to the relations above (replacing Tk by gk), then Gn =
Sn- Hint: Induct on n using cosets of the subgroup generated by
11. Fix a partition A and fix an ordering of standard A-tableaux ^ 1 ,^ 2 , —
Define the axial distance from A: to A; H- 1 in tableau U to be
Si = 5i{k,k + l) = {c'
where c, c' and r, r' are the column and row coordinates of k and A: -f 1,
respectively, in t { . Young ’s seminormal form assigns to each transposi-
tion r = (A:, A; + 1) the matrix px{r) with entries
Px(T)i,j =
l/6i if i = j,
1 — 1/5? if Tti = tj and i < j,
1 if rti = tj and i > j,
0 otherwise.
(a) Show that every row and column of px{r) has at most two nonzero
entries.
(b) Show that px can be extended to a representation of 5n, where
A h 72, by using the Coxeter relations of Exercise 10.
(c) Show that this representation is equivalent to the one afforded by
12. All matrices for this problem have rows and columns indexed by parti-
tions of 72 in dual lexicographic order. Define
A = (|<Sa n K^\) and B = {\S^\ • Kx^).
(Be sure to distinguish between the conjugacy class and the Kostka
number Kx^-) Show that A,B are upper triangular matrices and that
C = B{A^)~^ is the character table of Sn (with the columns listed in
reverse order). Hint: Use Frobenius reciprocity.
Use this method to calculate the character table for 54 .
2 .12. EXERCISES
89
13. Prove that, up to sign, the determinant of the character table for is
nn^.
Ahn A^GA
14. Prove the following results in two ways: once using representations and
once combinatorially.
(a) If Kxij, ^ 0, then X> fi.
(b) Suppose 11 and u are compositions with the same parts (only rear-
ranged). Then for any A, K\^ = K\i^. Hint: For the combinatorial
proof, consider the case where fi and n differ by an adjacent trans-
position of parts.
15. Let G be a group and let H < G have index two. Prove the following.
(a) H is normal in G.
(b) Every conjugacy class of G having nonempty intersection with H
becomes a conjugacy class of H or splits into two conjugacy classes
of H having equal size. Furthermore, the conjugacy class K oi G
does not split in H if and only if some k £ K commutes with some
g^H.
(c) Let X be an irreducible character of G. Then xIh is irreducible or
is the sum of two inequivalent irreducibles. Furthermore, xIh is
irreducible if and only if x{ 9 ) 7^ 0 for some g ^ H,
16. Let An denote the alternating subgroup of Sn and consider tt G <Sn
having cycle type A = (Ai, A 2 , . . . , A/).
(a) Show that tt G if and only if n — / is even.
(b) Prove that the conjugacy classes of Sn that split in An are those
where all parts of A are odd and distinct.
17. Use the previous two exercises and the character table of <S 4 to find the
character table of A 4 .
Chapter 3
Combinatorial Algorithms
Many results about representations of the symmetric group can be approached
in a purely combinatorial manner. The crucial link between these two view-
points is the fact (Theorem 2.6.5, part 2) that the number of standard Young
tableaux of given shape is the degree of the corresponding representation.
We begin this chapter with famed Robinson-Schensted algorithm [Rob 38,
Sch 61], which provides a bijective proof of the identity
Ahn
from part 3 of Theorem 2.6.5. This procedure has many surprising properties
that are surveyed in the sections that follow. This includes a discussion of
Schiitzenberger’s jeu de taquin [Scii 76], which is a crucial tool for many
results. In the last two sections we give a pair of formulae for
One is in terms of products, whereas the other involves determinants.
3.1 The Robinson-Schensted Algorithm
If we disregard its genesis, the identity
Ahn
can be regarded as a purely combinatorial statement. It says that the number
of elements in Sn is equal to the number of pairs of standard tableaux of
the same shape A as A varies over all partitions of n. Thus it should be
possible to give a purely combinatorial — i.e., bijective — proof of this formula.
The Robinson-Schensted correspondence does exactly that. It was originally
discovered by Robinson [Rob 38] and then found independently in quite a
91
92
CHAPTER 3. COMBINATORIAL ALGORITHMS
different form by Schensted [Sch 61]. It is the latter’s version of the algorithm
that we present.
The bijection is denoted by
7tM(P,Q),
where tt 6 Sn and P, Q are standard A-tableaux, A h n. We first describe the
map that, given a permutation, produces a pair of tableaux.
It— s
“tt — > (P, (5)” Suppose that tt is given in two-line notation as
1 2 ••• n
7T =
Xi X2 • • • Xn
We construct a sequence of tableaux pairs
(Po,Qo) = (0,0), {PuQll {P 2 .Q 2 ). (Pn,gn) = (P,Q), (3.1)
where xi , X 2 , . . . , are inserted into the P’s and 1, 2, . . . , n are placed in the
<9’s so that sh P^ = sh Qk for all k. The operations of insertion and placement
will now be described.
Let P be a partial tableau, i.e., an array with distinct entries whose rows
and columns increase. (So a partial tableau will be standard if its elements
are precisely {1,2,..., n}.) Also, let x be an element not in P. To row insert
X into P, we proceed as follows (where := means replacement).
RSI Set R := the first row of P.
RS2 While x is less than some element of row P, do
RSa Let y be the smallest element of R greater than x and replace
y by x in P (denoted by P 4- x).
RSb Set X := y and P := the next row down.
RS3 Now X is greater than every element of P, so place x at the end of
row P and stop.
To illustrate, suppose x = 3 and
P =
12 5 8
4 7
6
9
To follow the path of the insertion of x into P, we put elements that are
displaced (or bumped) during the insertion in boldface type:
1258^3 1238 1238
4 7 4 7 ^5 45
6 6 6
9 9 9
12 3 8
4 5
6 7
9
f- 7
3.1. THE ROBINSON-SCHENSTED ALGORITHM
93
If the result of row inserting x into P yields the tableau P', then write
r.(P) = F'.
Note that the insertion rules have been carefully written so that P' still has
increasing rows and columns.
Placement of an element in a tableau is even easier than insertion. Suppose
that Q is a partial tableau of shape ji and that (z, j) is an outer corner of fi.
If k is greater than every element of Q, then to place k in Q at cell
merely set Qij k. The restriction on k guarantees that the new array is
still a partial tableau. For example, if we take
Q
1 2 5
4 7
6
8
then placing A; = 9 in cell (z, j) = (2, 3) yields
1 2 5
479
6
At last we can describe how to build the sequence (3.1) from the permu-
tation
1 2 ••• n
7T =
Xi X2 • • • Xn
Start with a pair (Po?Qo) of empty tableaux. Assuming that {Pk-i^Qk-i)
has been constructed, define (Pk^Qk) by
Pk — (^/c— 1)5
Qk = place k into Qk-i at the cell (z,j) where the
insertion terminates.
Note that the definition of Qk ensures that shP^ = shQk for all k. We
call P = Pn the P -tableau, or insertion tableau, of tt and write P = P{7t).
Similarly, Q = Qn is called the Q -tableau, or recording tableau, and denoted
by Q == Q(7 t).
Now we consider an example of the complete algorithm. Boldface numbers
are used for the elements of the lower line of tt and hence also for the elements
of the Pk. Let
TT =
1 2 3 4 5 6 7
4 2 3 6 5 1 7 *
(3.2)
94
CHAPTER 3. COMBINATORIAL ALGORITHMS
Then the tableaux constructed by the algorithm are
Pk.
0.
4,
2,
4
23,
4
2 3 6,
4
2 3 5,
4 6
13 5,
2 6
4
13 5 7
2 6
4
= P,
Qk ■
0.
1,
1,
2
1 3,
2
1 3 4,
2
1 3 4,
2 5
1 3 4,
2 5
13 4 7
2 5
= Q.
6 6
Theorem 3.1.1 ([Rob 38, Sch 61]) The map
7T^(p,g)
So
The main theorem about this correspondence is as follows.
is a bijection between elements of Sn and pairs of standard tableaux of the
same shape A h n.
Proof. To show that we have a bijection, it suffices to create an inverse.
g
“(P, Q) — > 7t” We merely reverse the preceding algorithm step by step.
We begin by defining (Pn^Qn) = {P^Q)- Assuming that {Pk,Qk) has been
constructed, we will find Xk (the kth element of tt) and {Pk-i,Qk-i)- To
avoid double subscripting in what follows, we use Pij to stand for the (i, j)
entry of Pk>
Find the cell (z, j) containing fc in Since this is the largest element in
Qk, Pij must have been the last element to be displaced in the construction
of Pk. We can now use the following procedure to delete Pij from P. For
convenience, we assume the existence of an empty zeroth row above the first
row of Pk.
SRI Set X := P{j and erase Pij.
Set R := the {i — l)st row of P^.
SR2 While R is not the zeroth row of P^, do
SRa Let y be the largest element of R smaller than x and replace
y by X in R.
SRb Set X := y and R the next row up.
SR3 Now X has been removed from the first row, so set Xk := x.
It is easy to see that Pk-i is Pk after the deletion process just described
is complete and Qk-i is Qk with the k erased. Continuing in this way, we
eventually recover all the elements of tt in reverse order. ■
The Robinson-Schensted algorithm has many surprising and beautiful
properties. The rest of this chapter is devoted to discussing some of them.
3.2. COLUMN INSERTION
95
3.2 Column Insertion
Obviously, we can define column insertion of x into P by replacing row by
column everywhere in RSl-3. If column insertion of x into P yields P', we
write
c.(P) - P'.
It turns out that the column and row insertion operators commute. Before
we can prove this, however, we need a lemma about the insertion path. The
reader should be able to supply the details of the proof.
Lemma 3.2.1 Let P be a partial tableau with x ^ P. Suppose that during
the insertion rx{P) — P' , the elements bumped from cells
respectively. Then
1. X < x' < x" < - • ,
2 . f > j" > r > • • • ,
3. P'j < Pij for allij. m
Proposition 3.2.2 ([Sch 61]) For any partial tableau P and distinct ele-
ments x,y ^ P,
^2/^a:(P) ~ ^a:^2/(P)*
Proof. Let m be the maximum element in {x, 2 /} UP. Then m cannot displace
any entry during any of the insertions. The proof breaks into cases depending
on where m is located.
Case 1: y = m. (The case where x = mis similar.) Represent P schemat-
ically as
P =
(3.3)
Since y is maximal, Cy applied to either P or rx{P) merely inserts y at the
end of the first column. Let x be the last element to be bumped during the
insertion rx{P) and suppose it comes to rest in cell u. If u is in the first
column, then
Cyf'x{P) —
— '^xCy(P).
96
CHAPTER 3, COMBINATORIAL ALGORITHMS
If u is in any other column, then both CyVxiP) and TxCy{P) are equal to
X
m
Case 2: m E P , We induct on the number of entries in P. Let P be P
with the m erased. Then CyVxiP) C CyVxiP) and rxCy{P) C rxCy{P). But
CyVxiP) = rxCy{P) by induction. Thus Cyrx{P) and TxCy{P) agree everywhere
except possibly on the location of m.
To show that m occupies the same position in both tableaux, let x be the
last element displaced during rx{P)^ say into cell u. Similarly, define y and v
for the insertion Cy{P), We now have two subcases, depending on whether u
and V are equal or not.
Subcase 2a: u = v. Represent P as in diagram (3.3). Then rx{P) and
Cy{P) are represented, respectively, by
X
and
By the previous lemma, the first insertion displaces only elements in columns
(weakly) to the right ol u = u, and the second aflfects only those in rows
(weakly) below. Thus rx follows the same insertion path when applied to both
P and Cy{P) until x is bumped from the row above u. A similar statement
holds for Cy in P and rx{P). Thus if x < y (the case x > y being similar),
then we have
Cyrx{P)
X
V
rxCy{P).
Note that x and y must end up in the same column.
Now consider what happens to m. If m is not in cell u = v, then none
of the insertions displace it. So CyVxiP) and rxCy{P) are both equal to the
previous diagram with m added in its cell. If m occupies it, then both CyTxiP)
and rxCy{P) equal the preceding diagram with m added in the column just
to the right of u. (In both cases m is bumped there by y, and in the former
it is first bumped one row down by x.)
3.3. INCREASING AND DECREASING SUBSEQUENCES
97
Subcase 2b: u ^ v. Let P have shape A and let CyVx{P) = rxCy{P) have
shape A. Comparing rx{P) with Cyrx{P) and Cy{P) with TxCy{P)^ we see that
A U C A and A U {t’} C A. Thus
X — Xu {u, v}.
Since the insertion paths for rx{P) and Cy{P) may cross, x and y are not
necessarily the entries of cells u and v in CyVxiP) = TxCy{P). However, one
can still verify that u is filled by Vx and v hy ry^ whatever the order of the
insertions. Consideration of all possible places where m could occur in P now
completes the proof. The details are similar to the end of Subcase 2a and are
left to the reader. ■
As an application of this previous proposition, let us consider the reversal
of 7T, denoted by tt^; i.e., if n = X\X 2 • . -Xn, then tt’^ = XnXn-i . . .xi. The
P-tableaux of tt and are closely related.
Theorem 3.2.3 ([Sch 61]) If P{7 t) = P, then P{'ir'^) = P^, where t denotes
transposition.
Proof. We have
= Tx, • • (definition of P(7 t^))
= Xxi’ " Txr^-iCxrti^) (initial tableau is empty)
= Cx^Tx^ • • • ( 0 ) (Proposition 3.2.2)
= Cxr^Cxr^-i • • • Cxi (0) (induction)
= PL (definition of column insertion) ■
We will characterize the Q-tableau of using Schiitzenberger’s operation of
evacuation (Theorem 3.9.4).
If we represent the elements of Sn as permutation matrices, then the
dihedral group of the square (all refiections and rotations bringing a square
back onto itself) acts on them. For example, reversal is just reflection in
the vertical axis. It would be interesting to see what each of these changes
in the permutation does to the output tableaux of the Robinson-Schensted
algorithm. This theme motivates us in several of the following sections.
3.3 Increasing and Decreasing Subsequences
One of Schensted’s main motivations for constructing his insertion algorithm,
as the title of [Sch 61] suggests, was to study increasing and decreasing sub-
sequences of a given sequence tt G It turns out that these are intimately
connected with the tableau P(7 t).
Definition 3.3.1 Given tt = x\X 2 -->Xn ^ Sn, an increasing (respectively,
decreasing) subsequence of tt is
Xi^ < Xi.^ <’ - < Xi^
98
CHAPTER 3. COMBINATORIAL ALGORITHMS
(respectively, Xi^ > Xi^ > - • > Xi^) where ii < Z 2 < * * • < The integer k
is the length of the subsequence. ■
By way of illustration, take the permutation
7t = 423651 7. (3.4)
Then an increasing subsequence of tt of length 4 is 2 3 5 7, and a decreasing
subsequence of length 3 is 4 3 1. In fact, it is easy to check that these are the
longest increasing and decreasing subsequences of tt. On page 94 we found
the P-tableau of tt to be
13 5 7
P{7t)= 2 6 . (3.5)
4
Note that the length of the first row of P(7 t) is 4, whereas the length of the
first column is 3. This is not a coincidence, as the next theorem shows.
Theorem 3.3.2 ([Sch 61]) Consider tt e Sn- The length of the longest
increasing subsequence of tt is the length of the first row of P{tt). The length
of the longest decreasing subsequence of tt is the length of the first column of
P{ir).
Proof. Since reversing a permutation turns decreasing sequences into increas-
ing ones, the second half of the theorem follows from the first and Theorem
3.2.3. To prove the first half, we actually demonstrate a stronger result. In
what follows, Pk-i is the the tableau formed after A: — 1 insertions of the
Robinson-Schensted algorithm.
Lemma 3.3.3 If tt = X\X 2 • • - Xn and Xk enters Pk-i i^i column j, then the
longest increasing subsequence of tt ending in Xk has length j.
Proof. We induct on k. The result is trivial for A; == 1, so suppose it holds
for all values up to A: — 1.
First we need to show the existence of an increasing subsequence of length
j ending in Xk- Let y be the element of Pk-i in cell (1, j — 1). Then we have
y < Xk^ since Xk enters in column j. Also, by induction, there is an increasing
subsequence a of length j—1 ending in y. Thus crxk is the desired subsequence.
Now we must prove that there cannot be a longer increasing subsequence
ending in Xk> Suppose that such a subsequence exists and let Xi be the
element preceding Xk in that subsequence. Then, by induction, when Xi is
inserted, it enters in some column (weakly) to the right of column j. Thus
the element y in cell (1, j) of Pi satisfies
y <Xi < Xk.
But by part 3 of Lemma 3.2.1, the entries in a given position of a tableau
never increase with subsequent insertions. Thus the element in cell (1, j) of
3.4. THE KNUTH RELATIONS
99
Pk-i is smaller than Xk, contradicting the fact that Xk displaces it. This
finishes the proof of the lemma and hence of Theorem 3.3.2. ■
Note that the first row of P{tt) need not be an increasing subsequence of
7T even though it has the right length; compare (3.4) and (3.5). However, an
increasing subsequence of tt of maximum length can be reconstructed from
the Robinson-Schensted algorithm.
It turns out that an interpretation can be given to the lengths of the other
rows and columns of -P(7 t), and we will do this in Section 3.5. First, however,
we must develop an appropriate tool for the proof.
3.4 The Knuth Relations
Suppose we wish to prove a theorem about the elements of a set that is divided
into equivalence classes. Then a common way to accomplish this is to show
that
1. the theorem holds for a particular element of each equivalence class,
and
2. if the theorem holds for one element of an equivalence class, then it
holds for all elements of the class.
The set Sn has the following useful equivalence relation on it.
Definition 3.4.1 Two permutations 7t,(7 G are said to be P -equivalent^
p
written tt = cr, if P{'k) = P{(j). ■
For example, the equivalence classes in 5a are
{1 2 3}, {2 1 3, 2 3 1}, {1 3 2,3 1 2}
corresponding to the P-tableaux
{3 2 1},
1
2 .
3
respectively.
We can prove a strengthening of Schensted’s Theorem (Theorem 3.3.2)
using this equivalence relation and the proof technique just outlined. How-
ever, we also need an alternative description of P-equivalence. This is given
by the Knuth relations.
Definition 3.4.2 Suppose x < y < z. Then 7T,a e Sn differ by a Knuth
1
relation of the first kind, written tt = a, if
1. 7T = xi . . . yxz . . .Xn and a = x\ . . . yzx . . .Xn or vice versa.
2
They differ by a Knuth relation of the second kind, written tt = cr, if
2. TV — x\ . . . xzy . . .Xn and a = x\ . . . zxy . . .Xn or vice versa.
100
CHAPTER 3. COMBINATORIAL ALGORITHMS
K
The two permutations are Knuth equivalent^ written tt = cr, if there is a
sequence of permutations such that
i j I
7T = 7Ti = 7T2 = • • • = TT/c = Cr,
where G {1, 2}. ■
Returning to S 3 we see that the only nontrivial Knuth relations are
2 13^231 and 132^312.
Thus the Knuth equivalence classes and P-equivalence classes coincide. This
always happens.
Theorem 3.4.3 ([Knu 70]) If TT^a £ Sji, then
K P
7T = cr 7T = cr.
P
Proof. “=>” It suffices to prove that ir = a whenever tt and a differ by a
2 1
Knuth relation. In fact, the result for = follows from the one for = because
2 1
TT = cr tt’’ = (by definitions)
P{ 7 T^) = P{cr'^) (proved later)
P{ 7 vY = P{aY (Theorem 3.2.3)
Pin) = P{a).
1
Now assume tt = a. Keeping the notation of Definition 3.4.2, let P be the
tableau obtained by inserting the elements before y (which are the same in
both 7T and a). Thus it suffices to prove
rzTxTyiP) = Ta^TzVyiP).
Intuitively, what we will show is that the insertion path of y creates a “bar-
rier,” so that the paths for x and ^ lie to the left and right of this line,
respectively, no matter in which order they are inserted. Since these two
paths do not intersect, VzVx and VxVz have the same effect.
We induct on the number of rows of P. If P has no rows, i.e., P == 0, then
it is easy to verify that both sequences of insertions yield the tableau
X z
y
Now let P have r > 0 rows and consider P = Ty{P). Suppose y enters P in
column displacing element y' . The crucial facts about P are
1- P\,j ^y for all j < fc, and
3.4. THE KNUTH RELATIONS
101
2. Pi I > y' for all I > k.
If the insertion of x is performed next, then, since x < x must enter in
some column j with j < k. Furthermore, if x' is the element displaced, then
we must have x' < y' because of our first crucial fact. Applying Vz to rxVy{P)
forces z > y to come in at column /, where / > /c for the same reason. Also,
the element z' displaced satisfies z' > y' by crucial fact 2.
Now consider rxrzTy{P). Since the crucial facts continue to hold, z and x
must enter in columns strictly to the right and weakly to the left of column
/c, respectively. Because these two sets of columns are disjoint, the entrance
of one element does not affect the entrance of the other. Thus the first rows
of rzrxTy{P) and rxTzry^P) are equal. In addition, the elements displaced in
both cases are x',y\ and z', satisfying x' < y' < z' . Thus if P' denotes the
bottom r — 1 rows of P, then the rest of our two tableaux can be found by
computing rz>rx>ryf{P') and rx'rz>ry/{P'). These last two arrays are equal by
induction, so we are done with this half of the proof.
Before continuing with the other implication, we need to introduce some
more concepts.
Definition 3.4.4 If P is a tableau, then the row word of P is the permutation
7Tp = RiRi-i . . . Pi,
where Pi , . . . , P/ are the rows of P. ■
For example, if
13 5 7
P= 26
4
then
7Tp = 4 2 6 1 3 5 7.
The following lemma is easy to verify directly from the definitions.
Lemma 3.4.5 If P is a standard tableau, then
7Tp
R-S
The reader may have noticed that most of the definitions and theorems
above involving standard tableaux and permutations make equally good sense
when applied to partial tableaux and partial permutations, which are bijec-
tions 7T : K L between two sets of positive integers. It K = {k\ < k 2 <
• • • < km}, then we can write tt in two-line form as
k\ /u2 * * * km
l\ I2 ■ ■ ■
where li = 7r{ki). Insertion of the ki and placement of the U set up a bijection
R— S
7T — y (P, Q) with partial tableaux. Henceforth, we assume the more general
case whenever it suits us to do so. Now back to the proof of Theorem 3.4.3.
102
CHAPTER 3. COMBINATORIAL ALGORITHMS
By transitivity of equivalence relations, it suffices to show that
K
^ ^ ^
7T = TTp
whenever P(7t) = P.
We induct on n, the number of elements in tt. Let x be the last element of
7T, so that 7T = tt'x, where tt' is a sequence of n — 1 integers. Let P' = P{7t').
Then, by induction,
TT = TTp'.
So it suffices to prove that
K
TVpfX = TTp.
In fact, we will show that the Knuth relations used to transform ttp'X into
TTp essentially simulate the insertion of x into the tableau P' (which of course
yields the tableau P).
Let the rows of P' he R\, Ri, where jRi = piP 2 • • - Pk- If ^ enters P'
in column j, then
Thus
• <pj
:_1 < X
<pj<--
■ <pk-
Ri...
■R2P1 •
..Pj-lpj.
^•Pk-
iPkX
Ri...
R 2 P\ ■
■ ■Pj-lpj ■
• •Pk-
iXPk
Ri...
R2P1 ■
■■Pj-lpj ■
..XPk
-iPk
Ri...
R2P1 ■
..Pj-lPjXpj + i.
••Pk
Ri...
R2P1 ■
..PjPj-lXpj+i .
••Pk
^ Ri... R2PjPi . ..pj-ixpjpi ...pk.
Now the tail of our permutation is exactly the first row of P = rx{P')- Also,
the element bumped from the first row of P', namely pj, is poised at the end
of the sequence corresponding to R 2 . Thus we can model its insertion into
the second row just as we did with x and Pi. Continuing in this manner, we
eventually obtain the row word of P, completing the proof. ■
3.5 Subsequences Again
Curtis Greene [Gre 74] gave a nice generalization of Schensted’s result on
increasing and decreasing sequences (Theorem 3.3.2). It involves unions of
such sequences.
3.5. SUBSEQUENCES AGAIN
103
Definition 3.5.1 Let tt be a sequence. A subsequence cr of tt is k -increasing
if, as a set, it can be written as
C7 = ai t±) CT2 1+) • • • l+lcTfe,
where the cr^ are increasing subsequences of tt. If the a* are all decreasing,
then we say that a is k- decreasing. Let
ikM = the length of tt’s longest /^-increasing subsequence
and
dk{7T) = the length of tt’s longest /c-decreasing subsequence. ■
Notice that the case k = 1 corresponds to sequences that are merely increasing
or decreasing. Using our canonical example permutation (3.4),
7t = 423651 7,
we see that longest 1-, 2-, and 3-increasing subsequences are given by
and
and
2 3 5 7
423657 = 2357 W4 6
4236517 - 2357l±)46l±)l,
respectively. Thus
2i(7t)=4, Z2(7t)=6, 23(7t) = 7. (3.6)
Recall that P(7 t) in (3.5) had shape A = (4, 2, 1), so that
Ai =4, Ai -f- A2 = 6, Ai + A2 ~h A3 = 7. (^•'^)
Comparing the values in (3.6) and (3.7), the reader should see a pattern.
To talk about the situation with /c-decreasing subsequences conveniently,
we need some notation concerning the columns of A.
Definition 3.5.2 Let A be a Ferrers diagram; then the conjugate of X is
a'-(a;,a',...,au
where A' is the length of the ith column of A. Otherwise put. A' is just the
transpose of the diagram of A and so it is sometimes denoted by AL ■
As an example of the preceding definition, note that if
104
CHAPTER 3. COMBINATORIAL ALGORITHMS
• • • •
A = (4,2,1)- • •
then
• • •
y =A‘= I • =(3,2, 1,1).
Greene’s theorem is as follows.
Theorem 3.5.3 ([Gre 74]) Given tt G Sn, let shP(7r) — (Ai, A 2 , . . . , A/)
with conjugate (A'^, A 2 , . • . , A^). Then for any k,
ifc(7r) — Ai + A2 + • • • + Afc,
^fc(Tr) = A'l + A 2 H
Proof. By Theorem 3.2.3, it suffices to prove the statemer^t for /c-increasing
subsequences.
We use the proof technique detailed at the beginning of Section 3.4 and
the equivalence relation is the one studied there. Given an equivalence class
corresponding to a tableau P, use the permutation ttp as the special repre-
sentative of that class. So we must first prove the result for the row word.
By construction, the first k rows of P form a /^-increasing subsequence
of TTp, so ikiT^p) > Ai A 2 -f • • • + A/c. To show that the reverse inequality
holds, note that any /(^-increasing subsequence can intersect a given decreasing
subsequence in at most k elements. Since the columns of P partition ttp into
decreasing subsequences,
m k
u(7Tp) < ^min(A',fc) = ^Aj.
i=l
Now we must show that the theorem holds for all permutations in the
equivalence class of P. Given tt in this class. Theorem 3.4.3 guarantees the
existence of permutations tti, 7T2, . . . , such that
K K K
7r = 7Ti=:7r2 — •••— TTj— TTp,
where each tt^+i differs from tt^ by a Knuth relation. Since all the have the
same P-tableau, we need to prove only that they all have the same value for
i
ik. Thus it suffices to show that if tt = a for i — 1, 2, then iki^r) = We
1 2
will do the case of =, leaving = as an exercise for the reader.
Suppose 7T — xi . . . yxz . . .Xn and a = xi . . . yzx . . . To prove that
iki'^) ^ we need to show only that any /c-increasing subsequence of tt
has a corresponding /c-increasing subsequence in a of the same length. Let
tt' = 7Ti l+l 7T2 y • • • y TTfc
3.5. SUBSEQUENCES AGAIN
105
be the subsequence of n. If x and 2 : are not in the same tt^ for any z, then the
TVi are also increasing subsequences of cr, and we are done.
Now suppose that x and z are both in tti (the choice of which of the
component subsequences is immaterial). If y ^ tt', then let
a I = 7Ti with X replaced by y.
Since x < y < a\ is still increasing and
Cr' = (Ji (+) 7T2 y • • • 1+) TTfc
is a subsequence of a of the right length. Finally, suppose y e say z/ G 7T2
(note that we can not have y e tti). Let
7t[ = the subsequence of tti up to and including x,
TT^' = the subsequence consisting of the rest of tti ,
TV 2 = the subsequence of 7T2 up to and including y,
7T2 = the subsequence consisting of the rest of 7T2.
Note that tt^ = for z = 1, 2. Construct
(7 1 ~ and (72 ~ ^ 2^1 5
which are increasing because x < y < miuTr^' and y < z, respectively. Also,
<t' = CTi 1 +) (72 1+) 7T3 1 +) • • • 1+) 7T/fc
is a subsequence of a because x and z are no longer in the same component
subsequence. Since its length is correct, we have proved the desired inequality.
To show that Zfc(7r) > Zfc(c7) is even easier. Since x and z are out of order
in (7, they can never be in the same component of a A;-increasing subsequence.
Thus we are reduced to the first case, and this finishes the proof of the
theorem. ■
The fact that Zfc(7r) = Ai + A 2 H h A^ does not imply that we can find
a /c-increasing subsequence of maximum length
7t' = 7Ti 1 +) 7T2 1 +) • • • 1 +) TTfc
such that the length of tt^ is A^ for all z. As an example [Gre 74], consider
7t = 24795136 8 .
Then
1 3 4 6 8
P(7r)= 2 4 9
7
so Z2(7 t) = 5 -j- 3 = 8. There is only one 2-increasing subsequence of tt having
length 8, namely,
7t' = 2479 l±) 136 8.
The reader can check that it is impossible to represent tt' as the disjoint union
of two increasing subsequences of lengths 5 and 3.
106
CHAPTER 3. COMBINATORIAL ALGORITHMS
3.6 Viennot’s Geometric Construction
We now return to our study (begun at the end of Section 3.2) of the effects that
various changes in a permutation tt have on the pair (P(7 t), We prove a
remarkable theorem of Schiitzenberger [Scii 63] stating that taking the inverse
g
of the permutation merely interchanges the two tableaux; i.e., if tt — > (P, Q),
g
then 7T“^ — > (Q,P). Our primary tool is a beautiful geometric description
of the Robinson-Schensted correspondence due to Viennot [Vie 76].
Consider the first quadrant of the Cartesian plane. Given a permutation
7T = X 1 X 2 . . . Xn, represent Xi by a box with coordinates (i,Xi) (compare this
to the permutation matrix of tt). Using our running example permutation
from equation (3.4),
7T-423651 7.
we obtain the following figure:
Now imagine a light shining from the origin so that each box casts a
shadow with boundaries parallel to the coordinate axes. For example, the
shadow cast by the box at (4, 6) looks like this:
3.6. VIENNOT’S GEOMETRIC CONSTRUCTION
107
Similar figures result for the other boxes. Consider those points of the permu-
tation that are in the shadow of no other point, in our case (1,4), (2,2), and
(6, 1). The first shadow line, Li, is the boundary of the combined shadows of
these boxes. In the next figure, the appropriate line has been thickened:
Note that this is a broken line consisting of line segments and exactly one
horizontal and one vertical ray.
To form the second shadow line, L 2 , one removes the boxes on the first
shadow line and repeats this procedure.
Definition 3.6.1 Given a permutation displayed in the plane, we form its
shadow lines Li,L 2 ,... as follows. Assuming that Li, . . . , L{-i have been
108
CHAPTER 3. COMBINATORIAL ALGORITHMS
constructed, remove all boxes on these lines. Then L{ is the boundary of the
shadow of the remaining boxes. The x-coordinate of Li is
XL- = the x-coordinate of Lfs vertical ray,
and the y- coordinate is
yii the y-coordinate of Lfs horizontal ray.
The shadow lines make up the shadow diagram o/tt. ■
In our example there are four shadow lines, and their x- and y-coordinates
are shown above and to the left of the following shadow diagram, respectively.
Compare the coordinates of our shadow lines with the first rows of the
tableaux
1357 1347
P(7t) =26 and Q(7 t) =25
4 6
computed on page 94. It seems as if
Pi , 3 = VLj and Qi j = xl^
for all j. In fact, even more is true. The boxes on line Lj are precisely those
elements passing through the (1, j) cell during the construction of P, as the
next result shows.
Lemma 3.6.2 Let the shadow diagram of tt = X\X 2 • • - be constructed as
before. Suppose the vertical line x = k intersects i of the shadow lines. Let
yj be the y-coordinate of the lowest point of the intersection with Lj. Then
the first row of the Pk = P{x\ . . .Xk) is
Ri=yi V2 ••• Vi-
(3.9)
3.6. VIENNOTS GEOMETRIC CONSTRUCTION
109
Proof. Induct on k, the lemma being trivial for k = 0. Assume that the
result holds for the line x = k and consider x = k-\-\. There are two cases.
If
Xk+\ > Vi, (3.10)
then the box {k + l,x/c+i) starts a new shadow line. So none of the values
2/1 , . . . , 2/2 change, and we obtain a new intersection,
Ui+l — ^/e+1 •
But by (3.9) and (3.10), the {k + l)st intersection merely causes to sit
at the end of the first row without displacing any other element. Thus the
lemma continues to be true.
If, on the other hand.
Vi <■■■ < Vj-1 < Xk+1 <Vj < ■■■ <Vi, (3.11)
then {k + l,Xfc+i) is added to line Lj. Thus the lowest coordinate on Lj
becomes
y'j = Xk+i,
and all other 2 /- values stay the same. Furthermore, equations (3.9) and (3.11)
ensure that the first row of P^+i is
yi ... yj^i y'j ... yi.
This is precisely what is predicted by the shadow diagram. ■
It follows from the proof of the previous lemma that the shadow diagram
of 7T can be read left to right like a time-line recording the construction of
P(7 t). At the kth stage, the line x = k intersects one shadow line in a ray
or line segment and all the rest in single points. In terms of the first row
of Pfc, a ray corresponds to placing an element at the end, a line segment
corresponds to displacing an element, and the points correspond to elements
that are unchanged.
We can now prove that equation (3.8) always holds.
Corollary 3.6.3 ([Vie 76]) If the permutation tt has Robins on- Schensted
tableaux (P, Q) and shadow lines Lj, then, for all j,
P \,3 = VLj and Qij = xlj .
Proof. The statement for P is just the case k = n of Lemma 3.6.2.
As for Q, the entry k is added to Q in cell (1, j) when Xk is greater than
every element of the first row of Pk~i- But the previous lemma’s proof shows
that this happens precisely when the line x = k intersects shadow line Lj in
a vertical ray. In other words, 2 /Lj = k = Qij as desired. ■
How do we recover the rest of the the P- and Q-tableaux from the shadow
diagram of tt? Consider the northeast corners of the shadow lines. These
no
CHAPTER 3. COMBINATORIAL ALGORITHMS
are marked with a dot in the diagram on page 108. If such a corner has
coordinates then, by the proof of Lemma 3.6.2, x' must be displaced
from the first row of Pk-i by the insertion of Xk- So the dots correspond to
the elements inserted into the second row during the construction of P. Thus
we can get the rest of the two tableaux by iterating the shadow diagram
construction. In our example, the second and third rows come from the
thickened and dashed lines, respectively, of the following diagram:
Formally, we have the following definition.
Definition 3.6.4 The ith skeleton o/tt G <Sn, is defined inductively by
7 t(^) = 7t and
(i) ^2 ’ ' * kjYi
h h Im^
where (A^i, /i), . . . , {km, Im) are the northeast corners of the shadow diagram
of listed in lexicographic order. The shadow lines for are denoted
by . ■
The next theorem should be clear, given Corollary 3.6.3 and the discussion
surrounding Lemma 3.6.2.
Theorem 3.6.5 ([Vie 76]) Suppose tt {P,Q)- Then is a partial
permutation such that
ttW ^
where pb) (respectively, consists of the rows i and below of P (respec-
tively, Q). Furthermore,
^ — y and Qij — (*)
j 3
for all i,j. m
3.6. VIENNOT’S GEOMETRIC CONSTRUCTION
111
It is now trivial to demonstrate Schiitzenberger’s theorem.
Theorem 3.6.6 ([Scii 63]) If tt e Sn, then
P(7t“^) = Q(7t) and = P(7t).
Proof. Taking the inverse of a permutation corresponds to reflecting the
shadow diagram in the line y = x. The theorem now follows from Theorem
3.6.5. ■
As an application of Theorem 3.6.6 we can find those transpositions that,
when applied to tt G 5n, leave Q{ir) invariant. Dual to our definition of
P-equi valence is the following.
Definition 3.6.7 Two permutations 7t,(7 G 5^ are said to be Q-equivalent,
Q
written tt = cr, if Q{7t) = <3(cr). ■
For example,
Q{2 1 3) = Q(3 1 2) = 2 ^ and Q(1 3 2) = Q(2 3 1) = J ^ ,
SO
213i312 and 132^23 1. (3.12)
We also have a dual notion for the Knuth relations.
Definition 3.6.8 Permutations 7T,a e Sn differ by a dual Knuth relation of
1 *
the first kind, written tt = or, if for some k,
1. 7T = ...k + l...k...k-\-2... and cr = .../cH-2...A;...fc + l...
or vice versa.
2 *
They differ by a dual Knuth relation of the second kind, written tt = cr, if for
some k,
2. 7T = ...k...k-\-2...k-\-l... and a = ...k-{-l...k-\-2...k...
or vice versa.
The two permutations are dual Knuth equivalent, written tt = cr, if there is
a sequence of permutations such that
7T = 7Ti
i* j*
= 7r2 =
r
_ _
• * = TTfc — cr.
where i,j,...,l G {1,2}. ■
112
CHAPTER 3. COMBINATORIAL ALGORITHMS
Note that the only two nontrivial dual Knuth relations in S 3 are
1* 2*
213^312 and 132^23 1.
These correspond exactly to (3.12).
The following lemma is obvious from the definitions. In fact, the definition
of the dual Knuth relations was concocted precisely so that this result should
hold.
Lemma 3.6.9 Ifn^a G Sn, then
^ -1 -1
7T = (J 4==^^ 7T = a . ■
Now it is an easy matter to derive the dual version of Knuth’s theorem
about P-equivalence (Theorem 3.4.3).
Theorem 3.6.10 //7r,cr G Sn, then
K* Q
7 T = a 7 T = a.
Proof. We have the following string of equivalences:
K* K.
7 T = a 7 T~^ = a~^ (Lemma 3.6.9)
P{7T~^) = P{cr~^) (Theorem 3.4.3)
Q(7t) = Q{cr). (Theorem 3.6.6) ■
3.7 Schiitzenberger’s Jeu de Taquin
The jeu de taquin (or “teasing game” ) of Schiitzenberger [Scii 76] is a powerful
tool. It can be used to give alternative descriptions of both the P- and Q-
tableaux of the Robinson-Schensted algorithm (Theorems 3.7.7 and 3.9.4) as
well as the ordinary and dual Knuth relations (Theorems 3.7.8 and 3.8.8).
To get the full-strength version of these concepts, we must generalize to
skew tableaux.
Definition 3.7.1 If ^ C A as Ferrers diagrams, then the corresponding skew
diagram^ or skew shape^ is the set of cells
A//i = {c : c G A and c 0 p}.
A skew diagram is normal if /x = 0. ■
If A = (3, 3, 2, 1) and p = (2, 1, 1), then we have the skew diagram
A//i =
3.7. SCHUTZENBERGER’S JEU DE TAQUIN
113
Of course, normal shapes are the left-justified ones we have been considering
all along.
The definitions of skew tableaux, standard skew tableaux, and so on, are
all as expected. In particular, the definition of the row word of a tableau still
makes sense in this setting. Thus we can say that two skew partial tableaux
K
P, Q are Knuth equivalent^ written P = Q, if
K
TTp = 7 Tq.
Similar definitions hold for the other equivalence relations that we have intro-
duced. Note that if n = X\X2 • > > Xn^ then we can make tt into a skew tableau
by putting Xi in the cell (n — i -f 1, 2 ) for all i. This object is called the antidi-
agonal strip tableau associated with tt and is also denoted by tt. For example,
if TT = 3142 (a good approximation, albeit without the decimal point), then
K K
So TT = cr as permutations if and only if tt = cr as tableaux.
We now come to the definition of a jeu de taquin slide^ which is essential
to all that follows.
Definition 3.7.2 Given a partial tableau P of shape A//i, we perform a
forward slide on P from cell c as follows.
FI Pick c to be an inner corner of /i.
F2 While c is not an inner corner of A do
Fa If c = (^, j), then let c' be the cell of min{Pi+i ^ , P^j+i}.
Fb Slide Pc' into cell c and let c c'.
If only one of Pi^ij^ Pij^i exists in step Fa, then the maximum is taken to
be that single value. We denote the resulting tableau by j^(P). Similarly, a
backward slide on P from cell c produces a tableau jc{P) as follows
B1 Pick c to be an outer corner of A.
B2 While c is not an outer corner of /i do
Ba If c — (i, j), then let d be the cell of max{P^_i.j, Pij_i}.
Bb Slide Pc> into cell c and let c c'. ■
114
CHAPTER 3. COMBINATORIAL ALGORITHMS
By way of illustration, let
6 8
P= 2 4 5 9.
1 3 7
We let a dot indicate the position of the empty cell as we perform a forward
slide from c = (1,3).
1
Thus
• 6 8
2 4 5 9
3 7
4
7
f(P) =
4
2 5
3 7
6 8
9
A backward slide from c = (3, 4) looks like the following.
6
9
2
3
6
2 4 5
3*7
2 •
3 4
6 8
5 9 .
7
So
MP) =
1 3
6 8
5 9
7
Note that a slide is an invertible operation. Specifically, if c is a cell for
a forward slide on P and the cell vacated by the slide is d, then a backward
slide into d restores P. In symbols.
uf(P) = P.
Similarly,
fjd(p) - p.
if the roles of d and c are reversed.
Of course, we may want to make many slides in succession.
(3.13)
(3.14)
Definition 3.7.3 A sequence of cells (ci, C 2 , • . • , q) is a slide sequence for a
tableau P if we can legally form P = Pq, Pi^ . . . , Pi, where Pi is obtained
from Pi-i by performing a slide into cell c*. Partial tableaux P and Q are
equivalent, written P = Q, '\i Q can be obtained from P by some sequence of
slides. ■
This equivalence relation is the same as Knuth equivalence, as the next
series of results shows.
3,7. SCHUTZENBERGER’S JEU DE TAQUIN
115
Proposition 3.7.4 ([Scii 76]) If P^Q are standard skew tableaux, then
P^Q=^P^Q.
Proof. By induction, it suffices to prove the theorem when P and Q differ by
a single slide. In fact, if we call the operation in steps Fb or Rb of the slide
definition a move, then we need to demonstrate the result only when P and
Q differ by a move. (The row word of a tableau with a hole in it can still be
defined by merely ignoring the hole.)
The conclusion is trivial if the move is horizontal because then ttp = ttq.
If the move is vertical, then we can clearly restrict to the case where P and Q
have only two rows. So suppose that x is the element being moved and that
where Ri and Si (respectively, Rr and Sr) are the left (respectively, right)
portions of the two rows.
Now induct on the number of elements in P (or Q). If both tableaux
consist only of x, then we are done.
Now suppose \Rr\ > |5r|. Let y be the rightmost element of Rr and let
P',Q' be P,Q, respectively, with y removed. By our assumption P' and Q'
are still skew tableaux, so applying induction yields
K
7Tp = TTp/y = TTq/2/ = TTq.
The case 15/1 > |R/| is handled similarly.
Thus we are reduced to considering \Rr\ = |5r| and \Ri\ = |5/|. Say
Ri=xi...Xj, Rr =
Si = Zi . . . Zj , Sr =Wi... Wk,
By induction, we may assume that one of j or k is positive. We will handle
the situation where j > 0, leaving the other case to the reader. The following
simple lemma will prove convenient.
Lemma 3.7.5 Suppose ai < a 2 < • • • < Un-
K
1. If X < ai, then ai . . . a^x = aixa 2 . . ,an>
K
2. If X > On, then xa\ . . .On— a\ . . . a^-ixa^- ■
116
CHAPTER 3. COMBINATORIAL ALGORITHMS
Since the rows and columns of P increase, we have xi < zi and xi < wi
for all i as well as a;i < x. Thus
7Tp = zi,.. ZjWi . . . WkXi . . . Xnxyi ...yk
K
= Z\X\Z 2 . . . ZjW \ . . . WkX 2 . . . Xjxyi . . .yk (Lemma 3.7.5, part 1)
K
= z\XiZ 2 . . . ZjXWi . . . WkX 2 . . . Xjyi . . .yk (induction)
K
= zi . . . ZjXWi . . . WkXi . . . Xjyi . . .yk (Lemma 3.7.5, part 1)
= ttq. ■
Schiitzenberger’s teasing game can be described in the following manner.
Definition 3.7.6 Given a partial skew tableau P, we play jeu de taquin by
choosing an arbitrary slide sequence that brings P to normal shape and then
applying the slides. The resulting tableau is denoted by j(P). ■
It is not obvious at first blush that j{P) is well defined — i.e., independent
of the slide sequence. However, it turns out that we will always get the
Robinson-Schensted P-tableau for the row word of P.
Theorem 3.7.7 ([Scii 76]) If P is a partial skew tableau that is brought to
a normal tableau P' by slides, then P' is unique. In fact, P' is the insertion
tableau for np.
Proof. By the previous proposition, ttp = TTp/. Thus by Knuth’s theorem on
P-equivalence (Theorem 3.4.3), ttp and 7Tp/ have the same insertion tableau.
Finally, Lemma 3.4.5 tells us that the insertion tableau for TTp/ is just P'
itself. ■
We end this section by showing that equivalence and Knuth equivalence
are indeed equivalent.
Theorem 3.7.8 ([Scii 76]) Let P and Q be partial skew tableaux. Then
P^Q P^Q.
Proof. The only-if direction is Proposition 3.7.4. For the other implication,
K
note that since P = Q, their row words must have the same P-tableau (The-
orem 3.4.3 again). So by the previous theorem, j{P) = j(Q) = P', say. Thus
we can take P into Q by performing the slide sequence taking P to P' and
then the inverse of the sequence taking Q to P'. Hence P = Q. m
3.8 Dual Equivalence
In the last section we gave a characterization of Knuth equivalence in terms
of slides (Theorem 3.7.8). It would be nice to have a corresponding charac-
terization of dual Knuth equivalence, and this was done by Haiman [Hai 92].
3.8. DUAL EQUIVALENCE
117
This machinery will also be useful when we prove the Littlewood-Richardson
rule in Section 4.9. Before stating Haiman’s key definition, we prove a useful
result about normal tableaux.
Proposition 3.8.1 Let P and Q be standard with the same normal shape
A h n. Then P = Q.
Proof. Induct on n, the proposition being trivial for n < 2. When n > 3,
let c and d be the inner corner cells containing n in P and Q, respectively.
There are two cases, depending on the relative positions of c and d.
U c = d, then let P' (respectively, Q') be P (respectively, Q) with the n
K*
erased. Now P' and Q' have the same shape, so by induction 7Tp> = ttq/.
But then we can apply the same sequence of dual Knuth relations to ttp and
K*
7Tq, the presence of n being immaterial. Thus P = Q in this case.
If c 7 ^ d, then it suffices to show the existence of two dual Knuth-equivalent
tableaux P' and Q' with n in cells c and d, respectively. (Because then, by
what we have shown in the first case, it follows that
K* K* K*
TTp == 7Tp/ = 7 Tq/ = 7Tq,
and we are done.) Let e be a lowest rightmost cell among all the cells on the
boundary of A between c and d. (The boundary of A is the set of all cells at
the end of a row or column of A.) Schematically, we might have the situation
e
c
Now let
P' = n, Pi = n-2, Pi = n-1;
Qc = n — 1, Qe — 72 — 2, = n.
and place the numbers 1, 2, . . . , n — 3 anywhere as long as they are in the
K*
same cells in both P' and Q'. By construction, np = ttq. m
The definition of dual equivalence is as follows.
Definition 3.8.2 Partial skew tableaux P and Q are dual equivalent^ written
♦
P ^ Q, if whenever we apply the same slide sequence to both P and Q, we
get resultant tableaux of the same shape. ■
Note that the empty slide sequence can be applied to any tableau, so
*
P = Q implies that shP = shQ. The next result is proved directly from our
definitions.
118
CHAPTER 3. COMBINATORIAL ALGORITHMS
Lemma 3.8.3 Let P = Q. If applying the same sequence of slides to both
*
tableaux yields P' and Q' , then P' = Q' . m
One tableau may be used to determine a sequence of slides for another as
follows. Let P and Q have shapes yi/v and X/ respectively. Then the cells of
Q, taken in the order determined by Q’s entries, are a sequence of backward
slides on P. Let jq{P) denote the result of applying this slide sequence to
P. Also let V = vq{P) stand for the tableau formed by the sequence of cells
vacated during the construction of jg(P), i.e.,
Y- j = k if ( 2 , j) was vacated when filling the cell of /c G Q (3.15)
Displaying the elements of P as boldface and those of Q in normal type, we
can compute an example.
123 23 23 3
413,143, 43, 24 = jg(P);
2 2 11
= vq(P)-
With P and Q as before, the entries of P taken in reverse order define a
sequence of forward slides on Q. Performing this sequence gives a tableau de-
noted by j^{Q). The vacating tableau V = v^{Q) is still defined by equation
(3.15), with P in place of Q. The reader can check that using our exam-
ple tableaux we obtain j^{Q) = vq{P) and v^{Q) = 3 q{P)^ This always
happens, although it is not obvious. The interested reader should consult
[Hai 92].
Prom these definitions it is clear that we can generalize equations (3.13)
and (3.14). Letting V = v^{Q) and W = vq{P), we have
3^3q{P) = P and jvf{Q)=Q- (3.16)
To show that dual equivalence and dual Knuth equivalence are really the
same, we have to concentrate on small tableaux first.
Definition 3.8.4 A partial tableau P is miniature if P has exactly three
elements. ■
The miniature tableaux are used to model the dual Knuth relations of the
first and second kinds.
Proposition 3.8.5 Let P and Q be distinct miniature tableaux of the same
shape X/fjL and content. Then
P% Q P^Q.
3.8. DUAL EQUIVALENCE
119
Proof. Without loss of generality, let P and Q be standard.
By induction on the number of slides, it suffices to show the follow-
ing. Let c be a cell for a slide on P, Q and let P', Q' be the resultant tableaux.
Then we must have
shP' = shg' and P' ^ Q'. (3.17)
This is a tedious case-by-case verification. First, we must write down all the
skew shapes with 3 cells (up to those translations that do not affect slides
so the number of diagrams will be finite). Then we must find the possible
tableau pairs for each shape (there will be at most two pairs corresponding to
1 * 2 *
= and =). Finally, all possible slides must be tried on each pair. We leave the
details to the reader. However, we will do one of the cases as an illustration.
Suppose that A = (2, 1) and // = 0. Then the only pair of tableaux of this
shape is
P 1 2 ,^13
P = ^ and Q = 2
1 *
or vice versa, and P = Q. The results of the three possible slides on P, Q are
given in the following table, from which it is easy to verify that (3.17) holds.
c
(1,3) (2,2) (3,1)
P' :
1 2
3
2
1 3
2
1
3
g' :
1 3
2
1
2 3
3
1
2
i*
r\j
1 *
r\j
2* r
r\j no
Let N he 8i normal standard tableau of shape /i. So P' == j^{P)
*
and g' = j^{Q) are normal miniature tableaux. Now P = Q implies that
shP' = shg'. This hypothesis also guarantees that v^{P) = v^{Q) = V,
say. Applying equation (3.16),
jy{P')=Pj^Q=jy{Q'),
which gives P' ^ Q'. Thus P' and Q' are distinct miniature tableaux of the
same normal shape. The only possibility is, then.
1 3
2
}■
120
CHAPTER 3. COMBINATORIAL ALGORITHMS
Since P' = Q\ we have, by what was proved in the forward direction,
P = jv{P') — jv{Q') = Q- ■
To make it more convenient to talk about miniature subtableaux of a
larger tableau, we make the following definition.
Definition 3.8.6 Let P and Q be standard skew tableaux with
shP = /i/i/l-m and shQ — A//ihn.
Then PC Q denotes the tableau of shape \/v \~ m n such that
iPiJO) = l
0)c I ifcGA//i. ■
Using the P and Q on page 118, we have
PUQ
1 2 3
4 5 7.
6
We need one more lemma before the main theorem of this section.
Lemma 3.8.7 ([Hai 92]) Let V, W, P, and Q he standard skew tableaux
with
shU = ix/v, shP = shQ = A//i, shW = k/X.
Then
P^Q =^VUPUW ^VUQUW.
Proof. Consider what happens in performing a single forward slide on U U
PU W, say into cell c. Because of the relative order of the elements in the U,
P, and W portions of the tableau, the slide can be broken up into three parts.
First of all, the slide travels through U, creating a new tableau V' = joiV)
and vacating some inner corner d of //. Then P becomes P' = jd(P), vacating
cell e, and finally W is transformed into W' — je{W). Thus jc{V UPUW) =
U' U P' U W'.
Now perform the same slide on V U Q U IT. Tableau V is replaced by
*
jc(V) = vacating d. If Q' = jd{Q), then, since P = Q, we have shP' =
shQ'. So e is vacated as before, and W becomes IT'. Thus jc{V U Q U IT) =
V' UQ' U IT' with P' = Q' by Lemma 3.8.3.
Now the preceding also holds, mutatis mutandis, to backward slides.
Hence applying the same slide to both V U P U W and T U P U IT yields
tableaux of the same shape that still satisfy the hypotheses of the lemma. By
induction, we are done. ■
We can now show that Proposition 3.8.5 actually holds for all pairs of
tableaux.
3.9. EVACUATION
121
Theorem 3.8.8 ([Hai 92]) Let P and Q be standard tableaux of the same
shape X/fi. Then
K* *
P ^ Q P^Q.
Proof. “=>” We need to consider only the case where P and Q differ by
a single dual Knuth relation, say the first (the second is similar). Now Q is
obtained from P by switching A: + 1 and fc -h 2 for some k. So
P=-VUP'UW ^ndQ = VuQ'uW,
where P' and Q' are the miniature subtableaux of P and Q, respectively, that
1 * *
contain A:, A: + 1, and A; + 2. By hypothesis, P' = Q', which implies P' = Q'
by Proposition 3.8.5. But then the lemma just proved applies to show that
P^Q.
“<^=” Let tableau N be of normal shape fi. Let
P' = f{P)&nAQ' = f{Q).
Since P = Q, we have P' = Q' (Lemma 3.8.3) and v^{P) = v^{Q) = V for
some tableau V. Thus, in particular, shP' = shQ', so that P' and Q' are
dual Knuth equivalent by Proposition 3.8.1. Now, by definition, we have a
sequence of dual Knuth relations
P'
= Pk = Q',
where ...,/' G {1, 2}. Hence the proof of the forward direction of Propo-
sition 3.8.5 and equation (3.16) show that
P = jv{P') - 3 v{P2) = • • • = jv{Q') = Q
for some G {1, 2}. This finishes the proof of the theorem. ■
3.9 Evacuation
We now return to our project of determining the effect that a reflection or
rotation of the permutation matrix for tt has on the tableaux P{7t) and Q(7 t).
We have already seen what happens when tt is replaced by (Theorem
3.6.6). Also, Theorem 3.2.3 tells us what the P-tableau of looks like.
Since these two operations correspond to reflections that generate the dihedral
group of the square, we will be done as soon as we determine To do
this, another concept, called evacuation [Scii 63], is needed.
Definition 3.9.1 Let Q be a partial skew tableau and let m be the minimal
element of Q. Then the delta operator applied to Q yields a new tableau, AQ,
defined as follows.
122
CHAPTER 3. COMBINATORIAL ALGORITHMS
D1 Erase m from its cell, c, in Q.
D2 Perform the slide on the resultant tableau.
If Q is standard with n elements, then the evacuation tableau for' Q is the
vacating tableau V = evQ for the sequence
Q, AQ, A^Q, A"g.
That is,
Vd = n — i if cell d was vacated when passing
Taking
Q =
13 4 7
2 5
6
from A^Q to ■
we compute ev Q
as follows, using
dots as placeholders
for cells not yet filled.
1 3 4 7,
2 3 4 7, 3 4 7,
4 7, 5 7,
6 7, 7
A*g : 2 5
5
5
5 6
6
6
6
6
• • • •,
6, ••56, • • 5
6, ..SG, *256
ev Q : • •
• 7
• 7
• 7 • 7
3 7 3 7
•
• •
• 4
4 4
Completing the last slide we obtain
12 5 6
ev Q = 3 7
4
Note that Q was the Q-tableau of our example permutation 7t = 4236517
in (3.4). The reader who wishes to anticipate Theorem 3.9.4 should now
compute and compare the results.
Since the delta operator is defined in terms of slides, it is not surprising
that it commutes with jeu de taquin.
Lemma 3.9.2 Let Q he any skew partial tableau; then
jA{Q) = Aj{Q).
Proof. Let P be Q with its minimum element m erased from cell c. We write
this as P = Q — {m}. Then
jA{Q)=j{P)
by the definition of A and the uniqueness of the j operator (Theorem 3.7.7).
3.9. EVACUATION
123
We now show that any forward slide on Q can be mimicked by a slide on
P. Let Q' = j^{Q) for some cell d. There are two cases. If d is not vertically
or horizontally adjacent to c, then it is legal to form P' = j^{P). Since m is
involved in neither slide, we again have that P' = Q' — {m}.
If d is adjacent to c, then the first move of j^{Q) will put m in cell d
and then proceed as a slide into c. Thus letting P' = j^{P) will preserve the
relationship between P' and Q' as before.
Continuing in this manner, when we obtain j(Q), we will have a corre-
sponding P", which is just j{Q) with m erased from the (1, 1) cell. Thus
Aj(Q) = = j{P) = jA{Q). m
As a corollary, we obtain our first relationship between the jeu de taquin
and the Q-tableau of the Robinson-Schensted algorithm.
Proposition 3.9.3 ([Scii 63]) Suppose tt = xiX 2 • . - Xn G «Sn and let
_ _ 2 3 • • ■ n
~ X2 Xs • • • Xn '
Then
Q{Tf) = AQ(7r).
Proof. Consider <t = 7r“^ and a = 7f“^. Note that the lower line of a is
obtained from the lower line of a by deleting the minimum element, 1.
By Theorem 3.6.6, it suffices to show that
P{a) = AP(a).
If we view a and a as antidiagonal strip tableaux, then a = Aa. Hence by
Theorem 3.7.7 and Lemma 3.9.2,
P{a) = P{Aa) = AP(cr). ■
Finally, we can complete our characterization of the image of tt^ under
the Robinson-Schensted map.
Theorem 3.9.4 ([Scii 63]) IfneSn, then
Q{7t'^) = evQ{7rY.
Proof. Let tt, ^ be as in the previous proposition with
1 2 ••• n-1
TT =
^ n — 1 * * ‘ ^2
Induct on n. Now
Q{7T^)-{n} = Qiw^) = Xn---X 2 )
= evQ{7fY (induction)
= evQ(7r)* - {n}. (Proposition 3.9.3)
124
CHAPTER 3. COMBINATORIAL ALGORITHMS
Thus we need to show only that n occupies the same cell in both Q{'k^)
and eYQ{7ry. Let shQ(7r) = A and shQ(^) = A. By Theorems 3.1.1 and
3.2.3 we have shQ(7r^) = A^ and shQ(^’^) = x \ Hence
cell of n in Q(7 t’^) = = Xn • • • X 2 )
= (A/A)^
= (cell of n in evQ(7r))L (Proposition 3.9.3) ■
3.10 The Hook Formula
There is an amazingly simple product formula for /^, the number of standard
A-tableaux. It involves objects called hooks.
Definition 3.10.1 If = (z,j) is a node in the diagram of A, then it has
hook
H, = Hij = {(ij') : f>j}U{{i'j) : i' > i}
with corresponding hooklength
hy — hi j — ®
To illustrate, if A = (4^,3^, 1), then the dotted cells in
are the hook i/ 2,2 with hooklength /i 2,2 = 6.
It is now easy to state the hook formula of Frame, Robinson, and Thrall.
Theorem 3.10.2 (Hook Formula [FRT 54]) // A h n, then
yA ^
n(i,j)€A
Before proving this theorem, let us pause for an example and an anecdote.
Suppose we wish to calculate the number of standard Young tableaux of shape
A = (2, 2, 1) h 5. The hooklengths are given in the array
where hi^j is placed in cell (^, j). Thus
.(2,2,1) ^ ^ 5
^ 4 - 3 - 2 -12
3.10. THE HOOK FORMULA
125
This result can be verified by listing all possible tableaux:
12 12 13 13 14
3 4, 3 5, 2 4, 2 5, 2 5.
5 4 5 4 3
The tale of how the hook formula was born is an amusing one. One
Thursday in May of 1953, Robinson was visiting Frame at Michigan State
University. Discussing the work of Staal [Sta 50] (a student of Robinson),
Frame was led to conjecture the hook formula. At first Robinson could not
believe that such a simple formula existed, but after trying some examples he
became convinced, and together they proved the identity. On Saturday they
went to the University of Michigan, where Frame presented their new result
after a lecture by Robinson. This surprised Thrall, who was in the audience,
because he had just proved the same result on the same day!
There are many different bijective proofs of the hook formula. Franzblau
and Zeilberger [F-Z 82] were the first to come up with a (complicated) bijec-
tion. Later, Zeilberger [Zei 84] gave a bijective version of a probabilistic proof
of Greene, Nijenhuis, and Wilf [GNW 79] (see Exercise 17). But the map was
still fairly complex. Also, Remmel [Rem 82] used the Garsia-Milne involution
principle [G-M 81] to produce a bijection as a composition of maps. It was not
until the 1990s that a truly straightforward one-to-one correspondence was
found (even though the proof that it is correct is still somewhat involved),
and that is the one we will present here. The algorithm was originally out-
lined by Pak and Stoyanovskii [PS 92], and then a complete proof was given
by these two authors together with Novelli [NPS 97]. (A generalization of
this method can be found in the work of Krattent haler [Kra pr].)
To get the hook formula in a form suitable for bijective proof, we rewrite
it as
nl = f^ n
(*d)€A
So it suffices to find a bijection
(P,J),
where shT = shP = sh J = A, T is an arbitrary Young tableau, P is a
standard tableau, and J is an array such that the number of choices for the
entry Ji^j is hij. More specifically, define the arm and leg of Hij to be
Ai,j = : f > j} and = {(*', j) : i' > i},
respectively, with corresponding arm length and leg length
and
In the previous example with A = (4^, 3^, 1) we have
A2,2-{(2,3), (2,4)}, a/2,2 = 2;
i:^2,2 = {(3,2), (4,2), (5,2)}; //2,2 = 3.
126
CHAPTER 3. COMBINATORIAL ALGORITHMS
Note that hij = alij + llij + 1. So our requirement on the array J will be
(3.18)
and we will call J a hook tableau.
“T — > (P, J)” The basic idea behind this map is simple. We will use
a modified jeu de taquin to unscramble the numbers in T so that rows and
columns increase to form P. The hook tableau, J, will keep track of the
unscrambling process so that T can be reconstructed from P.
We first need to totally order the cells of A by defining
{i,j) < (^^/) if and only if j > j' or j = j' and i > i'.
Label the cells of A in the given order
Cl < C2 < . . . < Cn-
So, for example, if A = (3, 3, 2), then the ordering is
C8
C5
C2
c?
C4
Cl
C6
C3
Now if T is a A-tableau and c is a cell, we let T-^ (respectively, be the
tableau containing all cells b o^T with h < c (respectively, h < c). Continuing
our example, if
3 2 7 2 7
T = 6 1 8 , then T<ce = 18.
4 5 4 5
Given T, we will construct a sequence of pairs
(Ti,Ji) = (T,0), (T2,J2), (T3,J3), ...,(T,,Jn) = (P,J), (3.19)
where 0 is the tableau of all zeros and for all k we have that is standard,
and that Jk is a hook tableau which is zero outside >
If T is a tableau and c is a cell of T, then we perform a modified forward
slide to form a new tableau j^{T) as follows.
NPSl Pick c such that is standard.
NPS2 While T-^ is not standard do
NPSa If c = (i, j), then let d be the cell of minjPi+i j, T^j+i}.
NPSb Exchange Tc and Tc> and let c := c'.
It is easy to see that the algorithm will terminate since will eventually
become standard, in the worst case when c reaches an inner corner of A. We
3.10. THE HOOK FORMULA
127
call the sequence, p, of cells that c passes through the path of c. To illustrate,
if
T=:
9 6 15
8 2 3 7
4
and c = (1,2), then here is the computation of j^{T), where the moving
element is in boldface:
9615 9165 9135
T=8237, 8237, 8267 = f(T).
4 4 4
So in this case the path of c is
p = ((l,2), (1,3), (2,3)).
We can now construct the sequence (3.19). The initial pair (Ti, Ji) is
already defined. Assuming that we have (Tfc_i , Jk-i) satisfying the conditions
after (3.19), then let
Furthermore, if starts at Cfc = (i,j) and ends at then = Jk-i
except for the values
{Jk)h,j
( {Jk-i)h+i,j - 1 ioTi<h<i',
\ f - j for h - i'.
It is not hard to see that if Jk-i was a hook tableau, then Jk will still be one.
Starting with the tableau
6 2
T= 4 3 ,
5 1
here is the whole algorithm
6 :
Tfc : 4 :
6
4
5
1
4
5
= P,
Jk
0 0
0 -1
0 0
0 -2
0 0
0 0
0 -2
0 0
1 0
0 -2
1 0
1 0
Theorem 3.10.3 ([NPS 97]) For fixed X, the map
2
0
0
J.
T ® (P, J)
just defined is a bijection between tableaux T and pairs (P, J) with P a stan-
dard tableau and J a hook tableau such that shT = shP = sh J = A.
128
CHAPTER 3. COMBINATORIAL ALGORITHMS
Proof. As usual, we will create an inverse.
T” To determine the sequence (Tn, Jn)^ - • (Ti, Ji) we
start with (Tn,Jn) = Assuming that (Tk,Jk) has been constructed,
we first need to consider which cells could have been the end of the path for
j^'^(Tk-i). The set of candidate cells for Ck = (ioijo) in Tk is
— {(^ )j ) • 'I ^ ^05 j — jo T {Jk)i',jo^ i'^k)i'Jo ^ 0}*
So if we have
/ 13 18 10 1 4 0 0 0 -2 -2 \
17 15 2 3 5 0 0 -1 -2 0
- 14 16 6 7 11 ’ 0 0 1 1 0 ’
\ 19 12 89 0000
then the candidate cells for cio = (2,3) correspond to the nonnegative entries
of Jio in cells (i',3) with i' > 2, namely when (Jio) 3,3 == 1 and (Jio) 4,3 = 0*
So
Cio = {(3,3 + 1), (4,3 + 0)} -{(3,4), (4,3)}.
We must now modify backward slides as we did forward ones. Note that
the following algorithm is to be applied only to pairs {Tk,Jk) that occur as
output of the forward part of the algorithm. To avoid double subscripting,
we let (r',J')-(T,,J^).
SPNl Pick c eCk where Ck = («o? jo)-
SPN2 While c 7^ do
SPNa If c — («, j), then let c' be the cell of max{T/_i T/j_i} where
we use 0 in place of ^ if A: < 0 or / < jo-
SPNb Exchange T^ and T^, and let c c'. ■
It is not clear that this procedure is well defined, since it might be possible,
a priori, that we never have c — c^. However, we will prove that Ck is always
reached for any c E: Ck- We denote the resulting tableau by jc{Tk) and
note that the condition in SPNa guarantees that since T^^^ was standard,
jc{Tk)^^^ will be standard as well. The sequence of cells encountered during
the execution of jc will be called the reverse path r = rc corresponding to c,
and we let
Rk — {^c • ^ ^ Ck}-
Going back to the previous example, we have
j(3,4) (^10)
j(4,3) (^10)
13
18
10
1
4
17
15
7
3
5
14
16
2
6
11
19
12
8
9
13
18
10
1
4
17
15
8
3
5
14
16
2
7
11
19
12
6
9
r( 3 , 4 ) = ((3,4), (3,3), (2,3)};
r(4,3) = ((4,3), (3,3), (2,3)}.
3.10. THE HOOK FORMULA
129
To finish our description of the inverse algorithm, we must decide which of
the candidate cells to use for our modified backward slide. Define the code of
any reverse path r by replacing the steps of the form — by N (for
northward) and those of the form (2, j), j — 1) hy W (for westward), and
then reading the sequence of N's and VF’s backwards. Continuing with our
running example, the codes for ^(3^4) and r*(4^3) are NW and AT AT, respectively.
Define a lexicographic order on codes, and thus on reverse paths, by letting
AT < 0 < IT. (When comparing codes of different lengths, the shorter one is
padded out with 0’s.). So in our example NW > NN^ and so /'(3,4) > ^(4,3)-
Now given (T^, Jk) and Ck = we find the largest element r' in TZk with
corresponding candidate cell c' = (^^/) and let
Tk-i = jc'{Tk)-
Furthermore, Jk-i = Jk except for the values
{Jk-l)hj —
{
{Jk)h-i,j T 1
0
for i < h < z',
for h = i.
(3.20)
Unlike the forward definition, it is not obvious that Jk-i is well defined, since
we might have {Jk-i)h,j > GLh,j for some h. We will see shortly that this
cannot happen.
So our next order of business is to show that SPNl-2 and (3.20) are well
defined. First, however, we must set up some definitions and notation. Note
that we can use SPNl-2 to define a reverse path, r, for any candidate cell,
even if it does not pass through Ck = by letting r terminate when it
reaches the top of column jo (which SPN2a will force it to do if it does not
reach Ck). To talk about the relative positions of cells and paths, say that
cell (z, j) is west (respectively weakly west) of cell if z = z' and j < j'
(respectively, j < j'). Cell c is west of path p' if it is west of some cell d G p'.
Finally, path p is west of path p' if, for each c G p that is in the same row as a
cell of p', c is west of some d G p'. Similar definitions apply to north, weakly
north, etc.
We are now ready to start the well-definedness proofs, which will be by
reverse induction on fc. For ease of notation, let T' = Tjt, T" = T^-i, R! =
R'' z= Rk-\^ and so forth, and let Ck = (^o, Jo)- The following lemma is
fundamental for all that follows.
Lemma 3.10.4 Suppose that all reverse paths in R' go through (zq, jo)- Then
Tq with initial cell (zq, jg) G C' is the largest reverse path in R' if and only if
any initial cell {i',j') G C' of d G R\ r' 7^ r'o, satisfies
R1 Zq < z' < z'o and {i\j') is west and weakly south ofr^, or
R2 i' > Zq and d enters row Zg weakly west of r'o.
When reading the proof of this lemma, the reader may find it useful to refer
to the following diagram, which gives an example where A = (14^, 13^, 12),
(^0, jo) = (2, 2 ), (zq, jo) = (7, 11), the lines indicate the steps of Tq, black dots
are for possible cells in C', and white dots are for all other cells.
130
CHAPTER 3. COMBINATORIAL ALGORITHMS
oooooooooooooo
O • ♦ Q oooooooooo
©••ooooooooooo
o • • #—•—•—0 0000000
o • • • • • i #-■■■■# - o o
Proof. Since both r' and Tq end up at (io, Jo)? they must intersect somewhere
and coincide after their intersection. For the forward implication, assume that
neither R1 nor R2 hold. Then the only other possibilities force r' to join Vq
with a W step or start on Pq after an N step of Vq. In either case r' > Tq, a
contradiction.
For the other direction, if r' satisfies R1 or R2, then either r' C Tq and Tq
joins r' with a W step, or r' 2 ^0 joins Tq with an N step. This gives
r' < Vq. m
For our induction step we will need another lemma.
Lemma 3.10.5 If Vq goes through (to, jo) o^nd is north of some cell of a
reverse path r” G IZ" , then r" goes through {io + 1, jo)*
Proof. I claim that Tq is north of every cell of r" after the one given in the
hypothesis of the lemma. This forces r" through (zq-I-I, jo) as r' goes through
jo)* If the claim is false, then let (ii, ji) be the first cell of r" after the
given one that is northmost on Tq in its column. So the previous cell of r"
must have been (zi -f 1, ji). But, by construction of Tq,
rpf rriff ^ rjiH rpf
Ji “ — 1 ^ 1 ~ 1*
So r" should have moved from (ii + 1, ji) to ( 2*1 + 1, ji — 1) rather than (ii, ji),
a contradiction. ■
We will now simultaneously resolve both of our concerns about the inverse
algorithm being well defined.
Proposition 3.10.6 For all k, the hook tableau Jk is well defined and all
reverse paths go through Ck = (^ 0 ? jo) ^/e*
Proof. The proposition is true for io = 1 by definition of the algorithm. So
by induction, we can assume that the statement is true for J' and T' and
prove it for J" and T". By Lemma 3.10.4 and (3.20), we see that if r" G R"
starts at {i" ^j”) with io < i" < io, then (i", j") is south and weakly west
of Tq as in the following figure (which the reader should compare with the
previous one).
3.10. THE HOOK FORMULA
131
o
o
o o
o
o
o
o
o
o
o
o
o
o
o
o-o-c
o
o
o
o
o
o
o
o
o
o
o
•
• <
>
o
o
o
o
o
o
o
o
o
o
o
•
• J
h-
-o
-o
o
o
o
o
o
o
o
o
•
•
•
-o
-o
-o
-c
o
o
o
•
•
•
•
•
•
•
i
>
o
o
o
•
•
•
•
•
•
•
i
>
o
o
o
•
•
•
•
•
•
•
1
1
•
•
o
•
•
•
•
•
•
•
1
1
•
So, in particular, (i", j") G A and J" is well defined. Also, by Lemma 3.10.5,
these must go through {io -f 1, jo)- If reverse path
in T' starting at Then r" and r' must agree up to and including the
first cell on r' in a column weakly west of column Jq. Since this cell is south
of Tq, we are done by Lemma 3.10.5. ■
It remains to show that the two maps are inverses of each other. Actually,
it suffices to show that
1. if the pair (T", J") is derived from (P, 0) by A; — 1 applications of NPS,
then applying NPS followed by SPN is the identity map, and
2. if the pair (T', J') is derived from (T, J) by n — A: applications of SPN,
then applying SPN followed by NPS is the identity map.
The second statement is true because modified forward and backward slides
are step-by-step inverses, and there is no question about which cell to use to
undo the backward slide.
For the first one, we must show that if (T', J') is the pair after the last
application of NPS, then the path pg for fho forward slide is just the
largest reverse path G IV read backwards. We first need a lemma whose
statement resembles that of Lemma 3.10.4.
Lemma 3.10.7 Suppose that Pq ends at (^O)io) not consist solely
of E (east) steps. Then the initial cell {i” ^j") G C" of any r" G W satisfies
1. io -\-l <i" <i^ and {i",j") is south and weakly west of p^, or
2. i" > Zq and r" enters row ig weakly west of Pq. m
The proof of this result is by contradiction. It is also very similar to that of
Lemma 3.10.5, and so is left to the reader.
By Lemma 3.10.4, to show that pg backwards is the greatest element of
1Z\ it suffices to show the following.
Lemma 3.10.8 The initial cell {i\j') G C' of any r' G IV other than the
reverse path of Pq satisfies
1 . iQ <i' < Zg and (z', j') is west and weakly south o/pg, or
2 . i' > Zg and r' enters row Zg weakly west o/pg. ■
132
CHAPTER 3. COMBINATORIAL ALGORITHMS
The proof of this last lemma is similar to that of Proposition 3.10.6, with
Lemma 3.10.7 taking the place of Lemma 3.10.4. It, too, is left to the reader.
This completes the proof of Theorem 3.10.3. ■
3.11 The Determinantal Formula
The determinantal formula for is much older than the hook formula. In
fact, the former was known to Frobenius [Pro 00, Fro 03] and Young [You 02].
In the following theorem we set 1/r! = 0 if r < 0.
Theorem 3.11.1 (Determinantal Formula) ^ (Ai, A 2 , . . . , A/) h n, then
f^ = n\
{Xi — i + j)\
where the determinant is I by I .
To remember the denominators in the determinant, note that the parts of
A occur along the main diagonal. The other entries in a given row are found
by decreasing or increasing the number inside the factorial by 1 for every step
taken to the left or right, respectively. Applying this result to A = ( 2 , 2 , 1 ),
we get
y( 2 , 2 ,l) ^ 51 ,
1 / 2 !
1 / 1 !
0
1/3! 1/4!
1/2! 1/3!
1 / 0 ! 1 / 1 !
= 5 ,
which agrees with our computation in Section 3.10.
Proof (of the determinantal formula). It suffices to show that the
determinant yields the denominator of the hook formula. Prom our definitions
\i P I = hi^i P ij so
1
1
(Ai — i + j)\
(^2,1 — 1 + /)!
Since every row of this determinant is of the form
1 1 11 "
_(/i-/ + l)! ' ‘ ‘ {h-2)\ {h - 1 )! h\ _
and we will use only elementary column operations, we will display only the
first row in what follows:
1
1
1 1 1
(hi, 1 — 2 )! (hi , 1 — 1)1 hi,i!
(^ 2,1 — 1 + j)\
3.12. EXERCISES
133
i=i
hi,i(hi,i — l)---(hi^i—l+2) ••• /ii,i(^i,i — 1) /ii,i 1
n
^1,1 (^1,1 ••• /ii,i(^i,i — 1) ^1,1 — 1 1
hi, i(hi,i-l)-(hi, 1-1+2) ••• (hi,i-l)(hi,i-2) hi,i-l 1
= n
i=l
hi,i!
(hi.i-l)(hi,i-2)-(hi,i-l+l) ■■■ (hi,i-l)(hi,i-2) hi, 1-1 1
n
2=1
hi I
(hi, 1-3)1 (hi, 1-2)1 (hi,i-l)l
But by induction on n, this last determinant is just 1/ IIvga where
A = (Ai — 1, A 2 — 1, . . . , A/ — 1)
= A with its first column removed.
Note that A may no longer have I rows, even though the determinant is still
I X /. However, induction can still apply, since the portion of the determinant
corresponding to rows of A of length zero is upper triangular with ones on the
diagonal. Now A contains all the hooklengths hij in A for z > 2. Thus
1 'n,l — •f'v ^
^=1 VGA
as desired. ■
Notice that the same argument in reverse can be used to prove the hook
formula for from the determinantal one. This was Frame, Robinson, and
Thrall’s original method of proof [FRT 54, pages 317-318], except that they
started from a slightly different version of the Frobenius- Young formula (see
Exercise 20).
3.12 Exercises
1. Prove Lemma 3.2.1.
134
CHAPTER 3. COMBINATORIAL ALGORITHMS
2. Let P be a partial tableau with x,y ^ P. Suppose the paths of insertion
for rx{P) = P' and ry(P') = P" are
(l,ji),...,(r,jV) and (1, jJ), . . . , (r', j',),
respectively. Show the following.
(a) li X <y, then ji < j[ for all i < r' and r' < r; i.e., the path for ry
lies strictly to the right of the path for r^c.
(b) If X > y, then ji > j[ for alH < r and r' > r; i.e., the path for Vy
lies weakly to the left of the path for Tx>
3. (a) If the Robinson-Schensted algorithm is used only to find the length
of the longest increasing subsequence of tt G <Sn, find its computa-
tional complexity in the worst case.
(b) Use the algorithm to find all increasing subsequences of tt of max-
imum length.
(c) Use Viennot’s construction to find all increasing subsequences of
7T of maximum length.
4. (a) Use the results of this chapter to prove the Erdos-Szekeres theo-
rem: Given any tt G <Snm+i? then tt contains either an increasing
subsequence of length n + 1 or a decreasing subsequence of length
m + 1.
(b) This theorem can be made into a game as follows. Player A picks
xi G 5 = {1, 2, . . . , nmTl}. Then player B picks X 2 E. S with X 2
x\. The players continue to alternate picking distinct elements of
S until the sequence X\X2 • . • contains an increasing subsequence
of length n -h 1 or a decreasing subsequence of length m + 1. In
the achievement (respectively, avoidance) version of the game, the
last player to move wins (respectively, loses). With m arbitrary,
find winning strategies for n < 2 in the achievement game and
for n < 1 in the avoidance game. It is an open problem to find a
strategy in general.
5. Prove Lemma 3.4.5 and describe the Q-tableau of 7Tp.
6. Let TT = . . . Xn be a partial permutation. The Greene invariant of tt
is the sequence of increasing subsequence lengths
i(7r) = (ii(7r),i2(7r),...,i„(7r)).
If TT G vSn, then let
and
= X\ . . .Xj
7T(j) = the subsequence of tt containing all elements < j.
3.12. EXERCISES
135
(a) Show that P{7t) = P{cr) if and only if — ^(^0)) J
a direct argument.
(b) Show that Qin) = Q{cr) if and only if i{7r^^^) = for all j by
a direct argument.
(c) Show that (a) and (b) are equivalent statements.
7. Recall that an involution is a map tt from a set to itself such that tt^ is
the identity. Prove the following facts about involutions.
(a) A permutation tt is an involution if and only if P(7t) = Q{7r).
Thus there is a bijection between involutions in Sn and standard
tableaux with n elements.
(b) The number of fixedpoints in an involution tt is the number of
columns of odd length in P(7t).
(c) [Frobenius-Schur] The number of involutions in Sn is given by
(d) We have
l-3-5---(2n-l) = ^
Ah2n
A^even
where A' even means that the conjugate of A has only even parts.
8. Let TT = xiX 2 > . • Xn be any sequence of positive integers, possibly with
repetitions. Define the P -tableau o/tt, P(7t), to be
'f'Xn'^Xn-l * ■ ’ '^Xi ( 0)5
where the row insertion operator is defined as usual.
(a) Show that P{tt) is a semistandard tableau.
(b) Find and prove the semistandard analogues of each of the follow-
ing results in the text. Some of the definitions may have to be
modified. In each case you should try to obtain the new result as
a corollary of the old one rather than rewriting the old proof to
account for repetitions.
i. Theorem 3.4.3.
ii. Theorem 3.5.3.
iii. Theorem 3.7.7.
9. Let F be a standard shifted tableau. Let P^ denote the left-justified
semistandard tableau gotten by pasting together P and as was done
for shapes in Exercise 21.
Also define the insertion operation ix{P) by row inserting x into P until
an element comes to rest at the end of a row (and insertion terminates)
or a diagonal element in cell (d, d) is displaced. In the latter case Pd,d
bumps into column d+1, and bumping continues by columns until some
element comes to rest at the end of a column.
136
CHAPTER 3. COMBINATORIAL ALGORITHMS
(a) Show that (ixP)^ = Cxrx(P^), where the column insertion operator
is modified so that x displaces the element of the first column
greater than or equal to x, and so on.
(b) Find and prove the shifted analogues of each of the following results
in the text. Some of the definitions may have to be modified. In
each case you should try to obtain the new result as a corollary of
the old one rather than rewriting the old proof to account for the
shifted tableaux.
i. Theorem 3.4.3.
ii. Theorem 3.5.3.
hi. Theorem 3.7.7.
10. Reprove Proposition 3.8.1 using the Robinson-Schensted algorithm.
11. Consider the dihedral group D4. Then g £ D4 acts on tt G 5 ^ by
applying the symmetry of the square corresponding to g to the permu-
tation matrix corresponding to tt. For every g £ D 4 , describe P{g 7 r)
and Q{g 7 r) in terms of P(7r) and Q(7 t). Be sure your description is as
simple as possible.
12. Let A be a partially ordered set. A lower order ideal of P is L C P
such that X £ L implies that y G L for all y < x. Upper order ideals
U are defined by the reverse inequality. If P = L|+) [/, where L and U
are lower and upper order ideals, respectively, then define analogues of
the jeu de taquin and vacating tableau on natural labelings of P (see
Definition 4.2.5) such that
j^{U) = vu(L) and v^{U) = ju{L).
Hint: Show that
3^{U)\i^v^{U) = ju{L)\i^vu{L)
by expressing each slide as a composition of operators that switch ele-
ments i and z + 1 in a labeling as long as it remains natural.
13. Suppose TT = X 1 X 2 . . • is a permutation such that P = P(7r) has
rectangular shape. Let the complement of tt be
= 2/12/2 •••yn,
where i/j = n -f 1 — for all i. Also define the complement of a rect-
angular standard tableau P with n entries to be the array obtained by
replacing P^j with n + 1 — Pij for all (z, j) and then rotating the result
180®. Show that
p ( 7 t ") = {p^y.
3.12. EXERCISES
137
14. The reverse delta operator^ A', is the same as the delta operator of
Definition 3.9.1, with m being replaced by the maximal element of Q
and replaced by jc- The reverse evacuation tableau^ ev'Q, is the
vacating tableau for the sequence
Q, A'Q, A'2g, A'”Q.
Prove the following.
(a) Evacuation and reverse evacuation are involutions, i.e.,
eyevQ — ev' ev' Q = Q.
(b) If Q has rectangular shape, then
ev'evQ = Q.
15. Prove Lemma 3.10.7.
16. Prove Lemma 3.10.8.
17. Fix a partition A h n and consider the following algorithm due to
Greene, Nijenhuis, and Wilf.
GNWl Pick a node v e X with probability 1/n.
GNW2 While v is not an inner corner do
GNWa Pick a node v e Hy — {r>} with probability l/{hy — 1).
GNWb := v.
GNW3 Give the label n to the corner cell v that you have reached.
GNW4 Go back to step GNWl with A := A - {v} and n := n - 1,
repeating this outer loop until all cells of A are labeled.
The sequence of nodes generated by one pass through GN W 1-3 is called
a trial. In this exercise you will show that GNWl-4 produces any given
standard A-tableau P with probability
prob(P) = (3.21)
n!
(a) Show that the algorithm always terminates and produces a stan-
dard tableau.
(b) Let (a,/?) be the cell of P containing n and let prob(a,/3) be the
probability that the first trial ends there. Prove that (3.21) follows
by induction from the formula
prob(a, /5) = - n f 1 +
n fJi V hi,j3
. (3.22)
138
CHAPTER 3. COMBINATORIAL ALGORITHMS
(c) Given a trial ending at (a,/3), we let the horizontal projection of
the trial be
I = {i ^ a : V = (i,j) for some v on the trial}.
The vertical projection is defined analogously. Let prob/ j(a,/3)
denote the sum of the probabilities of all trials terminating at (a, /?)
with horizontal projection I and vertical projection J. Show that
and use this to prove (3.22). (Hint: Induct on |7U J|.)
(d) Give an alternative proof of the hook formula based on this exer-
cise.
18. A poset r is a rooted tree if it has a unique minimal element and its
Hasse diagram has no cycles. If G r, then define its hook to be
Hy = {w e r : w >v}
with corresponding hooklength hy = \Hy\.
(a) Show that
= n\
T
where the sum is over all r with n nodes.
(b) Show in three ways (inductively, probabilistically, and using jeu de
taquin) that if r has n nodes, then the number of natural labelings
of r (see Definition 4.2.5) is
19. A two-person ballot sequence is a permutation tc = X\X 2 ^ • X 2 n of n ones
and n twos such that, for any prefix tt^ — x\X 2 • • - the number of
ones in iTk is at least as great as the number of twos. The nth Catalan
number, C^ is the number of such sequences.
(a) Prove the recurrence
C'n+l = CnCo + Cn-lCi "!-•*• + CqCji
for n > 0.
(b) The Catalan numbers also count the following sets. Show this in
two ways: by verifying that the recurrence is satisfied and by giving
a bijection with a set of objects already known to be counted by
the Cn-
3.12. EXERCISES
139
i. Standard tableaux of shape (n,n).
ii. Permutations tt e Sn with longest decreasing subsequence of
length at most two.
iii. Sequences of positive integers
1 ^ Cil ^ CL2 ^ ‘ CLn
such that ai < i for all i.
iv. Binary trees on n nodes.
V. Pairings of 2n labeled points on a circle with chords that do
not cross.
vi. Triangulations of a convex (n 4- 2)-gon by diagonals.
vii. Lattice paths of the type considered in Section 4.5 from (0, 0)
to (n, n) that stay weakly below the diagonal y = x.
viii. Noncrossing partitions of {1, 2, . . . , n}, i.e., ways of writing
{1, 2, . . . ,n} == (+) B 2 y • • • l+J
such that a,c e Bi and b,d e Bj with a < b < c < d implies
i = j.
(c) Prove, by using results of this chapter, that
1 /2n
n + 1 V n
20. If A = (Ai, A 2 , . . . , A/) h n, then derive the following formulae
-X ^ - Aj - ?+j)
^ -* + /)!
_ ~ ^3 a)
21. A partition A = (Ai, A 2 , . . . , A/) h n is strict if Ai > A 2 > • • • > A/.
Given any strict partition, the associated shifted shape A* indents row
i of the normal shape so that it starts on the diagonal square (z, i). For
example, the shifted shape of A = (4, 3, 1) is
A*
• •
The shifted hook of (z, j) G A* is
= : f > j} 'J {{i' , j) ■■ *' > *} U {0' + 1,/) : j' > j + 1}
with shifted hooklength /i*^ = For A* as before, the shifted
hooklengths are
7 5 4 2
4 3 1.
1
140
CHAPTER 3. COMBINATORIAL ALGORITHMS
(a) Show that the can be obtained as ordinary hooklengths in
a left-justified diagram obtained by pasting together A* with its
transpose.
(b) Defining standard shifted tableaux in the obvious way, show that
the number of such arrays is given by
/
A*
Hi
n! , 1
Hot
ni<,(Ai + A,)^ (Xi-i+jy.
n!
Chapter 4
Symmetric Functions
We have seen how some results about representations of Sn can be proved
either by using general facts from representation theory or combinatorially.
There is a third approach using symmetric functions, which is our focus in
this chapter.
After giving some general background on formal power series, we derive
the hook generating function for semistandard tableaux (generalizing the
hook formula. Theorem 3.10.2). The method of proof is a beautiful algo-
rithmic bijection due to Hillman and Grassl [H-G 76].
Next, the symmetric functions themselves are introduced along with the
all-important Schur functions, s\. The Jacobi-Trudi determinants give al-
ternative expressions for s\ analogous to the determinant al form for and
lattice-path techniques can be used to provide a combinatorial proof. Other
definitions of the Schur function as a quotient of alternates or as the cycle in-
dicator for the irreducible characters of Sn are presented. The latter brings in
the characteristic map, which is an isomorphism between the algebra of sym-
metric functions and the algebra of class functions on the symmetric group
(which has the irreducible characters as a basis). Knuth’s generalization of
the Robinson-Schensted map [Knu 70] completes our survey of generating
function analogues for results from the previous chapter.
We end by coming full circle with two applications of symmetric functions
to representation theory. The first is the Littlewood- Richardson rule [L-R 34],
which decomposes a tensor product into irreducibles by looking at the cor-
responding product of Schur functions. We present a proof based on the jeu
de taquin and dual equivalence. The second is a theorem of Murnaghan-
Nakayama [Nak 40, Mur 37] which gives an algorithm for computing the ir-
reducible characters. Its proof involves much of the machinery that has been
introduced previously.
Those who would like a more extensive and algebraic treatment of sym-
metric functions should consult Macdonald’s book [Mac 79].
141
142
CHAPTER 4. SYMMETRIC FUNCTIONS
4.1 Introduction to Generating Functions
We start with the most basic definition.
Definition 4.1.1 Given a sequence (an)n>o = aQ,ai,a 2 , ... of complex num-
bers, the corresponding generating function is the power series
fix) = "Y^anX^.
n>0
If the On enumerate some set of combinatorial objects, then we say that f{x)
is the generating function for those objects. We also write
[x^]f{x) — the coefficient of x'^ in f{x) — On. ■
For example, if
def
On = the number of n-element subsets of {1, 2, 3} ==
then
ao = 1, ai = 3, 02 =3, as = 1, and an = 0 for n > 4,
so
f{x) = = l + 3x^ + = (1 +
is the generating function for subsets of {1, 2, 3}. Equivalently,
ni+x)^
It may seem surprising, but to obtain information about a sequence it is
often easier to manipulate its generating function. In particular, as we will see
shortly, sometimes there is no known simple expression for On and yet f{x)
is easy to compute. Extensive discussions of generating function techniques
can be found in the texts of Goulden and Jackson [G-J 83], Stanley [Stn 97,
Stn 99], and Wilf [Wil 90].
Note that all our power series are members of the formal power series ring
C[N] — : On G C for all n},
n>0
which is a ring under the usual operations of addition and multiplication. The
adjective formal refers to the fact that convergence questions are immaterial,
since we will never substitute a value for x. The variable and its powers are
merely being used to keep track of the coefficients.
We need techniques for deriving generating functions. The basic counting
rules for sets state that the English words or and and are the equivalent of
the mathematical operations + and x . Formally, we have the following.
4,1. INTRODUCTION TO GENERATING FUNCTIONS
143
Proposition 4.1.2 Let S and T be finite sets.
1. // 5 n T = 0, then
\S\ST\ = \S\ + \T\.
2. If S and T are arbitrary, then
\SxT\ = |5| • \T\. m
This result has an analogue for generating functions (see Proposition 4.1.6).
First, however, let us see how these rules can be used informally to compute
a few examples.
A basic method for finding the generating function for a given sequence
{O'n)n >0 is as follows:
1. Find a set S with a parameter such that the number of elements of S
whose parameter equals n is a„.
2. Express the elements of S in terms of or, and, and the parameter.
3. Translate this expression into a generating function using +, x, and
x^.
In our previous example, = ( ^ ) , so we can take
S = all subsets T of {1, 2, 3}.
The parameter that will produce the sequence is
n = n{T) = the number of elements in T.
Now we can express any such subset as
T=(10TorlGT) and (2 ^ T or 2 G T) and (3 ^ T or 3 G T).
Finally, translate this expression into a generating function. Remember that
the n in x'^ is the number of elements in T, so the statements z 0 T and z G T
become x^ and x^ , respectively. So
f{x) = {x^ + x^) . {x^ + x^) • {x^ + x^) = (1 + xf
as before.
For a more substantial example of the method, let us find the generating
function
^p(n)a;",
n>0
where p{n) is the number of partitions of n. Here, S is all partitions A =
(imi ^ ^ ^ _ 1^1 ig Qf parts. We have
A = ( 1 ^ G A or 1 ^ G A or G A or • • •)
and ( 2 ° G A or 2 ^ G A or 2 ^ G A or • • •)
and ( 3 ° G A or 3 ^ G A or 3 ^ G A or • • •) and • • • ,
144
CHAPTER 4. SYMMETRIC FUNCTIONS
which translates as
f{x) = (a;°+a:^+a;^+^+- • ■){x°+x^+x^'^‘^+- ■ -)(a;°+x^+a;^+^+- ••)•••• (4.1)
Thus we have proved a famous theorem of Euler [Eul 48].
Theorem 4.1.3 The generating function for partitions is
^p(n)a;"
n>0
1 1 1
1 — xl — x^^l — X^’
(4.2)
Several remarks about this result are in order. Despite the simplicity of
this generating function, there is no known closed-form formula for p{n) itself.
(However, there is an expression for p{n) as a sum due to Hardy, Ramanujan,
and Rademacher; see Theorem 5.1 on page 69 of Andrews [And 76].) This
illustrates the power of our approach.
Also, the reader should be suspicious of infinite products such as (4.2).
Are they really well-defined elements of C[[a:]]? To see what can go wrong,
try to find the coefficient of x in n^i(l "h ^)- To deal with this problem,
we need some definitions. Consider /(x), /i(x), /2(x), . . . G C[[x]]. Then we
write Yli>i fi{^) — /(^) product converges to f{x) if, for
every n,
N
[x'^]f{x) = [x"j JJ/i(a;)
i=l
whenever N is sufficiently large. Of course, how large N needs to be depends
on n.
A convenient condition for convergence is expressed in terms of the degree
of f{x) G C[[x]], where
degf{x) = smallest n such that x'^ has nonzero coefficient in f{x).
For example,
deg{x^ x^ + x^ -i ) = 2 .
The following proposition is not hard to prove and is left to the reader.
Proposition 4.1.4 If fi{x) G C[[x]] for i > I and limi_>oc deg(/i(x) — 1) ==
OO; then ni>i fi{^) converges, m
Note that this shows that the right-hand side of equation (4.2) makes
sense, since there deg{ fi{x) — 1) = i.
By carefully examining the derivation of Euler’s result, we see that the
term 1/(1 — x^) in the product counts the occurrences of i in the partition
A. Thus we can automatically construct other generating functions. For
example, if we let
Po(n) = the number of A h n with all parts odd.
4.1. INTRODUCTION TO GENERATING FUNCTIONS
145
then
n>0 2>1
We can also keep track of the number of parts. To illustrate, consider
Pd(^) = the number of A h n with all parts distinct,
i.e., no part of A appears more than once. Thus the only possibilities for
a part i are G A or G A. This amounts to cutting off the generating
function in (4.1) after the first two terms, so
^Pd(n)x” = ]^(1 +x*).
n>0 i>l
As a final demonstration of the utility of generating functions, we use
them to derive another theorem of Euler [Eul 48] .
Theorem 4.1.5 For all n, Pd(n) = Po(^)-
Proof. It suffices to show that Pd(n) and Poiji) have the same generating
function. But
11 ( 1 +^*)
i>\
n<i+-‘)ii
i>l i>\
1-X^
I —
n
i>l
n
1 — X^
1
It is high time to make more rigorous the steps used to derive generating
functions. The crucial definition is as follows. Let S' be a se. Then a weighting
o/ S is a function
wt : S — > C[[x]].
If s G S, then we usually let wt s = for some n, what we were calling a
parameter earlier. The associated weight generating function is
fs{x) = ^wts.
ses
It is a well-defined element of C[[a:]] as long as the number of s G S with
degwts = n is finite for every n. To redo our example for the partition
function, let S be the set of all partitions, and if A G S, then define
wt A =
146
CHAPTER 4. SYMMETRIC FUNCTIONS
This gives the weight generating function
fs{x) = XI
xes
= EE-”
n>0 Ahn
n>0
which is exactly what we wish to evaluate.
In order to manipulate weights, we need the corresponding and-or rules.
Proposition 4.1.6 Let S and T he weighted sets.
1. //5nT = 0, then
fsturix) = fs{x) + frix).
2. Let S and T be arbitrary and weight S xT by wt(s, t) = wts wt t. Then
fsxT = fs{x) ■ frix).
Proof. 1. If 5 and T do not intersect, then
fsti)T{x) = X
sG5l±lT
= ^ wt 5 + ^ wt 5
ses sgt
= fs{x)-^ frix).
2. For any two sets 5, T, we have
/sxt(x) = X wt(s,t)
{s,t)eSxT
= wt 5 wt t
sGS
teT
= ^ wt s ^ wt t
ses teT
= fs{x)Mx). m
Under suitable convergence conditions, this result can be extended to
infinite sums and products. Returning to the partition example:
5 = = : mi>0}
= (1^, l\ 1^, . . .) II (2°, 2\ 2^, . . .) II • • •
= ({1°} W {1^ w {i^} y • •’ ■) ii ({2°} W {2^} W {2^} w • . •) u • • • ,
where II rather than x is being used, since one is allowed to take only a finite
number of components such that rui ^ 0. So
fs(x) = (/{lO} + /{li} + /{12} H )(/{20} + /{2>} + /{22} H ) • • • •
Since f{x}(x) = wt A = we recover (4.1) as desired.
4.2. THE HILLMAN-GRASSL ALGORITHM
147
4.2 The Hillman- Grassl Algorithm
Just as we were able to enumerate standard A-tableaux in Chapter 3, we
wish to count the semistandard variety of given shape. However, since there
are now an infinite number of such arrays, we have to use generating func-
tions. One approach is to sum up the parts, as we do with partitions. This
leads to a beautiful hook generating function that was first discovered by
Stanley [Stn 71]. The combinatorial proof given next is due to Hillman and
Grassl [H-G 76].
Fix a shape A and let ssx{n) be the number of semistandard A-tableaux
T such that Xl (2 ~ Since all entries in row i of T are at least of
size z, we can replace each Ti^j by Tij — i to obtain a new type of tableau.
Definition 4.2.1 A reverse plane partition of shape A, T, is an array ob-
tained by replacing the nodes of A by nonnegative integers so that the rows
and columns are weakly increasing. If the entries in of T sum to n, we say
that T is a reverse plane partition of n. Let
rppx{n) = the number of reverse plane partitions of n having shape A. ■
The use of the term reverse for plane partitions where the parts increase
is a historical accident stemming from the fact that ordinary partitions are
usually written in weakly decreasing order.
From the definitions, we clearly have
ssA(n)a;” = ^ rppx{n)x"',
n>0 n>0
where m(A) = Thus it suffices to find the generating function for
reverse plane partitions. Once again, the hooklengths come into play.
Theorem 4.2.2 Fix a partition A. Then
Y,rppx{n)x^= n
n>0 {i,j)eX
Proof. By the discussion after Proposition 4.1.4, the coefficient of x'^ in this
product counts partitions of n, where each part is of the form hij for some
ihj) ^ (Note that the part hij is associated with the node (i,j) G A, so
parts hij and hk^i are considered different if (i, j) ^ (fc,/) even if hij = /i^,/
as integers.) To show that this coefficient equals the number of reverse plane
partitions T of n, it suflices to find a bijection
that is weight preserving, i.e.,
{i,j)e\ k
148
CHAPTER 4. SYMMETRIC FUNCTIONS
‘T -> 5 • • •)” Given T, we will produce a sequence of reverse
plane partitions
T = To, Ti , T 2 , . . . , T/ = tableau of zeros,
where will be obtained from Tk-i by subtracting 1 from all elements of a
certain path of cells pk in Tj^. Since we will always have \pk\ = for some
this will ensure the weight-preserving condition.
Define the path p = pi in T inductively as follows.
HGl Start p at (a, 6), the most northeast cell of T containing a nonzero entry.
HG2 Continue by
if (i, j) E p then
{
{i,j -l)ep
+ ep
if
otherwise.
In other words, move south unless forced to move west in order not to
violate the weakly increasing condition along the rows (once the ones
are subtracted).
HG3 Terminate p when the preceding induction rule fails. At this point we
must be at the end of some column, say column c.
It is easy to see that after subtracting 1 from the elements in p, the array
remains a reverse plane partition and the amount subtracted is ha^c-
As an example, let
12 2 2
3 3 3
3
Then (a, 6) = (1,4), and the path p is indicated by the dotted cells in the
following diagram:
After subtraction, we have
1111
Ti = 2 2 3
2
and = hi^i. To obtain the rest of the Tk, we iterate this process. The
complete list for our example, together with the corresponding is
1222
1111
0 0 0 0
0 00 0
00 0 0
0 0 0 0
Tk-. 3 3 3 ,
2 2 3 ,
12 3,
12 2,
111 ,
00 0 ;
3
2
1
1
1
0
^l,l5
^2,3,
h2,2,
h2,i-
4.2. THE HILLMAN-GRASSL ALGORITHM
149
Thus T
“(/iii ji,/iz 2 j 25 • • •) Given a partition of hooklengths, we must re-
build the reverse plane partition. First, however, we must know in what order
the hooklengths were removed.
Lemma 4.2.3 In the decomposition of T into hooklengths, the hooklength
hi'jf was removed before hi/f^j/f if and only if
> i', or i” == and j" < j'. (4.3)
Proof. Since (4.3) is a total order on the nodes of the shape, we need only
prove the only-if direction. By transitivity, it suffices to consider the case
where hif/jff is removed directly after hi'j'.
Let T' and T" be the arrays from which hi^^jf and were removed
using paths p' and p", respectively. By the choice of initial points and the
fact that entries decrease in passing from T' to T", we have i" > i' .
If i" > z', we are done. Otherwise, i" — i' and p" starts in a column weakly
to the west of p'. We claim that in this case p" can never pass through a node
strictly to the east of a node of p', forcing j” < j'. If not, then there is some
(s, t) e p' n p" such that (s, t — 1) e p' and (s l,t) G p". But the fact that
p' moved west implies T^ t — Since this equality continues to hold in
T" after the ones have been subtracted, p" is forced to move west as well, a
contradiction. ■
Returning to the construction of T, if we are given a partition of hook-
lengths, then order them as in the lemma; suppose we get
We then construct a sequence of tableaux, starting with the all-zero array.
^/-i5 • • • ? To = T,
by adding back the for A: = /, / — 1, . . . , 1. To add ha^c to T, we
construct a reverse path r along which to add ones.
GHl Start r at the most southern node in column c.
GH2 Continue by
if (z, j) G r then
{
{i,j + 1) G r
€ r
if Tij+i = Tij,
otherwise.
GH3 Terminate r when it passes through the east most node of the row a.
It is clear that this is a step-by-step inverse of the construction of the
path p. However, it is not clear that r is well defined, i.e., that it must pass
through the eastern end of row a. Thus to finish the proof of Theorem 4.2.2,
it suffices to prove a last lemma.
150
CHAPTER 4. SYMMETRIC FUNCTIONS
Lemma 4.2.4 Ifvk is the reverse path for then (i^, G rk-
Proof. Use reverse induction on k. The result is obvious when k = f by the
first alternative in step GH2.
For k < f , let r' = rk and r" = taj+i- Similarly, define T' ^T" ^ hi ' ^ and
hiffjn. By our ordering of the hooklengths, i' < i". If i' < z", then row i' of
T consists solely of zeros, and we are done as in the base case.
If i' = i'f then j/ ^ j// pf starts weakly to the east of p". By the
same arguments as in Lemma 4.3, p stays to the east of p'. Since p' reaches
the east end of row i' = i by assumption, so must p. m
It is natural to ask whether there is any relation between the hook formula
and the hook generating function. In fact, the former is a corollary of the
latter if we appeal to some general results of Stanley [Stn 71] about poset
partitions.
Definition 4.2.5 Let (A, <) be a partially ordered set. A reverse A-partition
of m is an order-preserving map
OL I A — V {0, 1 , 2 ,.. .}
such that YlveA ol{v) —m. If | A| = n, then an order-preserving bijection
/^ : A — {1, 2, . . . , n}
is called a natural labeling of A. ■
To see the connection with tableaux, partially order the cells of A in the
natural way,
(i,j) < <=> i < i' and j < f.
Then a natural labeling of A is just a standard A- tableau, whereas a reverse
A-partition is a reverse plane partition of shape A.
One of Stanley’s theorems about poset partitions is the following.
Theorem 4.2.6 ([Stn 71]) Let A be a poset with \A\ = n. Then the gener-
ating function for reverse A-partitions is
(1 — x)(l — • • • (1 — x'^) ’
where P{x) is a polynomial such that P{1) is the number of natural labelings
of A. m
In the case where A = A, we can compare this result with Theorem 4.2.2
and obtain
^ TT 1
(1 — x){l — • • • (1 — X^) 1 - *
4.3. THE RING OF SYMMETRIC FUNCTIONS
151
Thus
=
4.3 The Ring of Symmetric Functions
In order to keep track of more information with our generating functions, we
can use more than one variable. The ring of symmetric functions then arises
as a set of power series invariant under the action of all the symmetric groups.
Let X = {xi,X2,X3, . . .} be an infinite set of variables and consider the
formal power series ring C[[x]]. The monomial xf^xf^ • • -xf^ is said to have
degree n \i n = We also say that /(x) G C[[x]] is homogeneous of
degree n if every monomial in /(x) has degree n.
, For every n, there is a natural action of tt G *Sn on /(x) G C[[x]], namely,
7rf{xi,X2,X3,...) = f{x^l,XTr2,X^s,...), ( 4 . 4 )
where iri = i ior i > n. The simplest functions fixed by this action are gotten
by symmetrizing a monomial.
Definition 4.3.1 Let A = (Ai, A2, . . . , A/) be a partition. The monomial
symmetric function corresponding to A is
mx = mx{x) = ■ • • x^^ ,
where the sum is over all distinct monomials having exponents Ai, . . . , A/. ■
For example.
p(i)
lim
(1 -
n
1 n(t,j)eA
n\
^(2,1) = xjx2 + 0:1X2 + X1X3 + X1X3 + X2X3 4- X2X3 H .
Clearly, if A h n, then m^(x) is homogeneous of degree n.
Now we can define the symmetric functions that interest us.
Definition 4.3.2 The ring of symmetric functions is
A = A(x) = CmA,
i.e., the vector space spanned by all the m\. m
Note that A is really a ring, not just a vector space, since it is closed under
product. However, there are certain elements of C[[x]] invariant under (4.4)
that are not in A, such as riz>i(l + x^), which cannot be written as a finite
linear combination of mx.
152
CHAPTER 4. SYMMETRIC FUNCTIONS
We have the decomposition
A =
where A^ is the space spanned by all m\ of degree n. In fact, this is a grading
of A, since
f e and g e implies fg e (4.5)
Since the m\ are independent, we have the following result.
Proposition 4.3.3 The space has basis
{m\ : A h n}
and so has dimension p{n) , the number of partitions of n. ■
There are several other bases for A’^ that are of interest. To construct
them, we need the following families of symmetric functions.
Definition 4.3.4 The nth power sum symmetric function is
Pn = rri(^n) =
i>l
The nth elementary symmetric function is
Cfi — ^ ^ ^ii * * * ^in •
The nth complete homogeneous symmetric function is
hn — ^ m\ — ^ ^ ^ii ' * ' ^in
Al 7T ii • • • *^in.
As examples, when n = 3,
P3 = x\ + xl + xl + ---,
63 = XiX2X2,-\- XiX 2 X 4 ,-\- X\X^X 4 ,-]r X 2 X^X 4 ^-\ ,
^3 = x\-\- x\-\ h x\x2 + X1X2 H h X1X2X3 + X1X2X4 H .
The elementary function is just the sum of all square-free monomials
of degree n. As such, it can be considered as a weight generating function for
partitions with n distinct parts. Specifically, let
S = {\ : /(A)-n},
where l{\) is the number of parts of A, known as its length. If A = (Ai >
A 2 > • • • > An), we use the weight
wt A = xx^x\2 • •
which yields
6n(x) = fs{x).
Similarly, hn is the sum of all monomials of degree n and is the weight gen-
erating function for all partitions with n parts. What if we want to count
partitions with any number of parts?
4.3. THE RING OF SYMMETRIC FUNCTIONS
153
Proposition 4.3.5 We have the following generating functions
E{t) ^ e„(x)t” = JJ(1 + Xit),
H{t) 1^' Y. ^ n
n>0 i>\
Proof. Work in the ring C[[x, ^]]. For the elementary symmetric functions,
consider the set 5 = {A : A with distinct parts} with weight
wt'A = t'WwtA,
where wt is as before. Then
/s(x,t) = ^wt'A
Ag5
n>OZ(A)=n
= ^e„(x)t".
To obtain the product, write
s = ({1°} w {li}) X ({2°} W {2i}) X ({3°} W {3^}) x • • • ,
SO that
/^(x, t) = (1 -f Xit){l -f X 2 t){l + Xst) • • • .
The proof for the complete symmetric functions is analogous. ■
While we are computing generating functions, we might as well give one
for the power sums. Actually, it is easier to produce one for pn{x.)/n.
Proposition 4.3.6 We have the following generating function:
x:p.(x)^=inn(Y^.
n>l i>l ^ ^ ^
Proof. Using the Taylor expansion of In we obtain
A (1 - Xit)
(1 - Xit)
= EE
i>l n>l
= E^E<
n>l 2>1
154
CHAPTER 4. SYMMETRIC FUNCTIONS
In order to have enough elements for a basis of A^, we must have one
function for each partition of n according to Proposition 4.3.3. To extend
Definition 4.3.4 to A = (Ai, A 2 , . . . , A/), let
f X fxi f X 2 ' ' ' f Xi 1
where f ~ p, or h. We say that these functions are multiplicative. To
illustrate, if A = (2, 1), then
P(2,l) - [x\ + x\ x\ ){xi +X 2 +X^-\ ).
Theorem 4.3.7 The following are bases for A'^.
1. {px : Ahn}.
2. {ex ’ A h n}.
3. [hx : A h n}.
Proof.
1. Let C = (cA/i) be the matrix expressing the px in terms of the basis
m^. If we can find an ordering of partitions such that C is triangular
with nonzero entries down the diagonal, then C~^ exists, and the px
are also a basis. It turns out that lexicographic order will work. In fact,
we claim that
P\ = cxxmx + ^ (4.6)
Ait>A
where caa ^ 0. (This is actually stronger than our claim about C by
Proposition 2.2.6.) But if • • -x^"^ appears in
Px = (xj * + X 2 ' + • • -)(x^^ + X 2 " H ) • • •
then each /i^ must be a sum of Aj’s. Since adding together parts of a
partition makes it become larger in dominance order, mx must be the
smallest term that occurs.
2. In a similar manner we can show that there exist scalars dx^ such that
e\' =mx-\-'Y^dxnm^,
/iCA
where A' is the conjugate of A.
3. Since there are p{n) = dim functions /ia, it suffices to show that they
generate the basis e^. Since both sets of functions are multiplicative,
we may simply demonstrate that every is a polynomial in the hk>
From the products in Proposition 4.3.5, we see that
H{t)E{-t) = 1.
4.4. SCHUR FUNCTIONS
155
Substituting in the summations for H and E and picking out the coef-
ficient of t” on both sides yields
n
= 0
r=0
for n > 1. So
~ h\6n—\ ^2^n— 2 “1“ * * ’ j
which is a polynomial in the /I’s by induction on n. ■
Part 2 of this theorem is often called the fundamental theorem of symmet-
ric functions and stated as follows: Every symmetric function is a polynomial
in the elementary functions e^.
4.4 Schur Functions
There is a fifth basis for that is very important, the Schur functions. As
we will see, they are also intimately connected with the irreducible represen-
tations of Sn and tableaux. In fact, they are so protean that there are many
different ways to define them. In this section, we take the combinatorial
approach.
Given any composition /x = (/ii,/i 2 , . . . ,/i/), there is a corresponding
monomial weight in C[[x]]:
y/i
(4.7)
Now consider any generalized tableau T of shape A. It also has a weight,
namely,
= JJ , = x'^, (4.8)
bd)6A
where fj. is the content of T. For example, if
4 1
1 3
4
then
= X 1X3X4.
Definition 4.4.1 Given a partition A, the associated Schur function is
«a(x) = ^x^,
T
where the sum is over all semistandard A-tableaux T.
156
CHAPTER 4. SYMMETRIC FUNCTIONS
By way of illustration, if A = (2, 1), then some of the possible tableaux are
11 12 11 13 12 13 12 14
^*2 ’2 ’3 ’3 ’"*3 ’2 ’4 ’2
so
5(2,1) (x) = ^1^2 + Xixl + xlxs -f- Xixl H h 2XiX2Xs -h 2X1X20:4 H .
Note that if A = (n), then a one-rowed tableau is just a weakly increasing se-
quence of n positive integers, i.e., a partition with n parts (written backward),
so
S(„)(x) = /i„(x). (4.9)
If we have only one column, then the entries must increase from top to bottom,
so the partition must have distinct parts and thus
S( 1 ") = e„(x).
(4.10)
Finally, if A h n is arbitrary, then
[a;ia;2---a;„]sA(x) =
since pulling out this coefficient merely considers the standard tableaux.
Before we can show that the s\ are a basis for A^, we must verify that
they are indeed symmetric functions. We give two proofs of this fact, one
based on our results from representation theory and one combinatorial (the
latter being due to Knuth [Knu 70]).
Proposition 4.4.2 The function sx{x) is symmetric.
Proof 1. By definition of the Schur functions and Kostka numbers.
( 4 . 11 )
At
where the sum is over all compositions of n. Thus it is enough to show that
Kx^ - Kxn (4.12)
for any rearrangement /I of /i. But in this case and M^ are isomorphic
modules. Thus they have the same decomposition into irreducibles, and (4.12)
follows from Young’s rule (Theorem 2.11.2).
Proof 2. It suffices to show that
{i,i + 1)sa(x) = sa(x)
for each adjacent transposition. To this end, we describe an involution on
semistandard A- tableaux
T— >T'
4.4. SCHUR FUNCTIONS
157
such that the numbers of z’s and (z + l)’s are exchanged when passing from
T to T' (with all other multiplicities staying the same).
Given T, each column contains either an z, z-f 1 pair; exactly one of z, z-f 1;
or neither. Call the pairs fixed and all other occurrences of z or z + 1 free.
In each row switch the number of free z’s and (z + l)’s; i.e., if the the row
consists of k free z’s followed by I free (z H- l)’s then replace them by I free z’s
followed by k free (z + l)’s. To illustrate, if z = 2 and
1 1 1 1 2 2 2 2 2 3
T- 2 2 3 3 3 3
3
then the twos and threes in columns 2 through 4 and 7 through 10 are free.
So
1 1 1 1 2 2 2 3 3 3
r' =222333
3
The new tableau T' is still semistandard by the definition of free. Since
the fixed z’s and (z + l)’s come in pairs, this map has the desired exchange
property. It is also clearly an involution. ■
Using the ideas in the proof of Theorem 4.3.7, part 1, the following result
guarantees that the s\ are a basis.
Proposition 4.4.3 We have
sa = ^ Kx^m^,
li<X
where the sum is over partitions p (rather than compositions) and K\\ — 1-
Proof. By equation (4.11) and the symmetry of the Schur functions, we have
where the sum is over all partitions /i. We can prove that
{
0
1
if A ^ /i,
if A = /z.
in two different ways.
One is to appeal again to Young’s rule and Corollary 2.4.7. The other is
combinatorial. If 7 ^ 0, then consider a A-tableau T of content /z. Since
T is column-strict, all occurrences of the numbers 1,2, ...,z are in rows 1
through z. This implies that for all z.
/zi + /Z2 + • • • + /Zi ^ Ai "h A2 + • • • + Aj,
i.e., /z < A. Furthermore, if A = /z, then by the same reasoning there is only
one tableau of shape and content A, namely, the one where row z contains all
occurrences of z. (Some authors call this tableau superstandard.) ■
Corollary 4.4.4 The set {s\ : Ah n} is a basis for A^. m
158
CHAPTER 4. SYMMETRIC FUNCTIONS
4.5 The Jacobi- Trudi Determinants
The determinantal formula (Theorem 3.11.1) calculated the number of stan-
dard tableaux, /^. Analogously, the Jacobi-Trudi determinants provide an-
other expression for s\ in terms of elementary and complete symmetric func-
tions. Jacobi [Jac 41] was the first to obtain this result, and his student
Trudi [Tru 64] subsequently simplified it.
We have already seen the special 1x1 case of these determinants in
equations (4.9) and (4.10). The general result is as follows. Any symmetric
function with a negative subscript is defined to be zero.
Theorem 4.5.1 (Jacobi-Trudi Determinants) Let A = (Ai, A 2 , . . . , A/).
We have
and
^A' — \^Xi—i+j I ?
where X' is the conjugate of X and both determinants are I x 1.
Proof. We prove this theorem using a method of Lindstrom [Lin 73] that
was independently discovered and exploited by Gessel [Ges um] and Gessel-
Viennot [G-V 85, G-V ip]. (See also Karlin [Kar 88].) The crucial insight is
that one can view both tableaux and determinants as lattice paths. Consider
the plane Z x Z of integer lattice points. We consider (possibly infinite) paths
in this plane
p = 5i, 52, S3, • • • 5
where each step Si is of unit length northward (N) or eastward (E). Such a
path is shown in the following figure.
58
P =
52
5l
54
53
55
57
56 ,
Label the eastward steps of p using one of two labelings. The e-labeling
assigns to each eastward Si the label
-^(s^) — i-
4.5. THE JACOBI-TRUDI DETERMINANTS
159
The h-labeling gives Si the label
L{si) — (the number of northward Sj preceding Si) + 1.
Intuitively, in the h-labeling all the eastward steps on the line through the
origin of p are labeled 1, all those on the line one unit above are labeled 2,
and so on. Labeling our example path with each of the two possibilities yields
the next pair of diagrams.
h-labeling
It is convenient to extend Z x Z by the addition of some points at infinity.
Specifically, for each x G Z, add a point (x,oo) above every point on the
vertical line with coordinate x. We assume that a path can reach (x, oo) only
by ending with an infinite number of consecutive northward steps along this
line. If p starts at a vertex u and ends at a vertex v (which may be a point
at infinity), then we write u A u.
There are two weightings of paths corresponding to the two labelings. If
p has only a finite number of eastward steps, define
xp = n ^L{si)
SiGp
= n
siep
and
160
CHAPTER 4. SYMMETRIC FUNCTIONS
where each product is taken over the eastward Si in p. Note that is always
square-free and can be any monomial. So we have
en(x) =
p
and
h„(x) =
P
where both sums are over all paths (a, 6) A (a -f n, oo) for any fixed initial
vertex (a, 6).
Just as all paths between one pair of points describes a lone elementary or
complete symmetric function, all /-tuples of paths between / pairs of points
will be used to model the /-fold products contained in the Jacobi-Trudi deter-
minants. Let ui,U 2 , • • • ,ui and t;i, • • • , be fixed sets of initial and final
vertices. Consider a family of paths V = (pi,P 25 • • • where, for each z.
Pi
Ui -4
for some n e Si. Give weight to a family by letting
x^ =
i
with a similar definition for x^. Also, define the sign ofV to be
(-1)^ = SgnTT.
For example, the family
V4
Vs V2 Vi
1
1
1
•
• •
i
i
•
•
•
1
1
* i
1
1
1
„_! i_
,
-1
j 1
: f
•
U4 U'S U2 Ui
has
-V 4 2
X = X 2 X 3 X 4
4.5. THE JACOBI-TRUDI DETERMINANTS
161
and sign
(-l)^=sgn(l,2,3)(4) = +l.
We now concentrate on proving
sx = \hx^-i+j\, (4.13)
the details for the elementary determinant being similar and left to the reader.
Given A, pick initial points
u^ = {l-i,0) (4.14)
and the final ones
Vi — {Xi — i 1, (X)). (4.15)
The preceding 4-tuple is an example for A = (2, 2, 2, 1). From the choice of
vertices,
hx,-i+j =
p
Uj—yVi
Thus the set of /-tuples V with permutation tt corresponds to the term in the
determinant obtained from the entries with coordinates
(7t1,1), (7t2,2), (tt/,/).
So
I^Ai-i+j(x)| = ^(-l)^x^, (4.16)
V
where the sum is over all families of paths with initial points and final points
given by the Uj and Vi.
Next we show that all the terms on the right side of equation (4.16) cancel
in pairs except for those corresponding to /-tuples of nonintersecting paths.
To do this, we need a weight-preserving involution
V < ^ > V'
such that iiV = V' (corresponding to the 1-cycles of ^), then V is noninter-
secting and \iV ^V' (corresponding to the 2-cycles), then (—1)^ = —(—1)^ .
The basic idea is that if V contains some intersections, then we will find two
uniquely defined intersecting paths and switch their final portions.
Definition 4.5.2 Given define lV = V\ where
1. if Pi n pj = 0 for all z, j, then V = 'P',
2. otherwise, find the smallest index i such that the path pi intersects some
other path. Let vq be the first (SW-most) intersection on pi and let pj
be the other path through vq. (If there is more than one, choose pj such
that j is minimal.) Now define
V' = V with pi,pj replaced by p',p' ,
where
p'.=Ui^vo^ v^j and p' = uj -4 vq ^ v-ni- ■
162
CHAPTER 4, SYMMETRIC FUNCTIONS
By way of illustration, we can apply this map to our 4-tuple:
V4
U4
I
I
1
^^3
^3
V 2
Vi
V 4
^3
V 2
Vl
• •
•
•
•
1
1
I
1
•
1
1
1
i
•
i
1
•
+
•
•
1
1
•
-1
i
L
• •••••>- -►
1
• 1
1 1
1
\ +
1
1
: — u ^
,
-i i-_-
_ ,
: 1 Vo
1
1
1
1
: 1 Vo
: *
•
•
1
: i
,
U2 Ui U4 U2 Ui
Our choice of pi, pj^ and vq makes it possible to reconstruct them after
applying so the map is invertible, with itself as inverse. The nonintersecting
families are fixed by definition. For the intersecting /-tuples, the sign clearly
changes in passing from V to V'. Weight is preserved due to the fact that all
paths start on the same horizontal axis. Thus the set of labels is unchanged
by L.
Because of the cancellation in (4.16), the only thing that remains is to
show that
V
where the sum is now over all /-tuples of nonintersecting paths. But by our
choice of initial and final vertices, “P = (pi,P 2 , • • • is nonintersecting only
if it corresponds to the identity permutation. Thus (—1)^ = 4-1 and Ui ^
for all i. There is a simple bijection between such families and semistandard
tableaux. Given P, merely use the /i-labels of the ith path, listed in increasing
order, for the ith row of a tableau T. For example.
4.6. OTHER DEFINITIONS OF THE SCHUR FUNCTION
163
Vs V2 Vl
v =
4
1
Us
becomes the tableau
1 2
T= 2 3
4
2
In view of equations (4.14) and (4.15), T has shape A. By definition
the rows of T weakly increase. Finally, the columns must strictly increase
because the nonintersecting condition and choice of initial vertices force the
jth eastward step of to be higher than the corresponding step on pi.
Construction of the inverse map is an easy exercise. This completes the proof
of (4.13). ■
4.6 Other Definitions of the Schur Function
We would be remiss if we did not give the definition that Schur originally used
[Scu 01] (as a quotient of alternants) for the functions that bear his name.
This definition actually goes back to Jacobi [Jac 41], but Schur was the first
to notice the connection with irreducible characters of Sn , which is our third
way of defining s\.
First, we have to restrict A to I variables, where I = 1{X). (Actually any
I > 1{X) will do for s\.) Specifically, let
A; = {/(a;i,...,a;(,0,0,...) : /(x) € A}.
We use the abbreviation /(xi, ^ 2 , . . . , a;;) for a typical element of A;. For
example,
P3{xi,X2,Xs,X 4) = xl+xl+xl + X 4 .
Thus Ai is just the set of functions in the polynomial ring C[xi,X 2 , . • • ,x/]
that are fixed by the action (4.4) of Si.
Similarly, we can consider skew- symmetric functions in C[xi,X 2 , . . . ,x/],
which are those satisfying
7t/ = (sgnTr)/
164
CHAPTER 4. SYMMETRIC FUNCTIONS
for all 7T e Si. Just as we can obtain symmetric functions by symmetrizing a
monomial, we can obtain skew-symmetric ones by skew-symmetrization. Let
fi = (//i, /i 2 , . . . , /U/) be any composition with monomial given by (4.7).
Define the corresponding alternant by
a^{xi,...,xi) = ^(sgn7r)7Tx''.
It is easy to verify that is skew-symmetric. From the definition of a
determinant, we can also write
\l<i,j<l *
(4.17)
For example.
^( 4 , 2 , 1 ) (^1 5 ^ 2 ) ^3) == X1X2XS H- X^X2X^ + X1X2XI — XiX2X^ — X1X2X3 — XiX^x"^
xj
xl
Xl
xj
xj
xj
xl
The most famous example of an alternant arises when we let the compo-
sition be
(1-1,1 -2,..., 1,0).
In this case
i-j
as = X-
is the Vandermonde determinant. It is well-known that we have the factor-
ization
as= JJ (xi-Xj). (4-18)
If A is any partition of length I with two equal parts, then = 0, since
determinant (4.17) has two equal columns. So it does no harm to restrict our
attention to partitions with distinct parts. These all have the form A + J,
where A is an arbitrary partition and addition of integer vectors is done
componentwise. Furthermore, if we set xi = Xj in any alternant, then we
have two equal rows. So a\^s is divisible by all the terms in product (4.18),
which makes a\j^s/cL6 a polynomial. In fact, it must be symmetric, being
the quotient of skew-symmetric polynomials. The amazing thing, at least
from our point of view, is that this is actually the symmetric function s\. To
demonstrate this we follow the proof in Macdonald [Mac 79, page 25]. For
1 < i ^ ^5 let Cn ^ denote the elementary symmetric function in the variables
xi, . . . , Xj-i^Xj^i^ . . . iXi {xj omitted). We have the following lemma.
Lemma 4.6.1 Let /a = {fii, iJ. 2 , . . . •> /J>i) be any composition. Consider the I xl
matrices
= (xf ), = (V-Hi) and E = ((-l)'-e|i\).
4.6. OTHER DEFINITIONS OF THE SCHUR FUNCTION
165
Then
= H^E.
Proof. Consider the generating function for the ,
n=0 i^j
We can now mimic the proof of Theorem 4.3.7, part 3. Since
we can extract the coefficient of on both sides. This yields
-k -
k=l
which is equivalent to what we wished to prove.
Corollary 4.6.2 Let X have length 1. Then
sa =
as
where all functions are polynomials in . . . ^xi.
Proof. Taking determinants in the lemma, we obtain
\A^\ = \H ^\ • \E\,
(4.19)
where |yl^| = a^. First of all, let fi = 6. In this case Hs = (hi-j), which
is upper unitriangular and thus has determinant 1. Plugging this into (4.19)
gives \E\ = as.
Now, letting p = X-\- S in the same equation, we have
0'X-\-S
as
= \Hx+s\ = \hx,-
i-\-j i
Hence we are done, by the Jacobi-Trudi theorem. ■
Our last description of the s\ will involve the characters of Sn. To see the
connection, let us reexamine the change-of-basis matrix between the mono-
mial and power sum symmetric functions introduced in (4.6). Let us compute
a small example. When n = 3, we obtain
P(3)
= xf + H
= m(3),
P(2,l)
= {x\ T ~f" • • ’)(^i H" ^2 T •
••) = "i(3) +^(2.1),
P(13)
= (a:^i X2 ‘
= m( 3 ) + 3m(2,i) + 6m(i3)
Comparing this with the character values 0^ of the permutation modules
in Example 2.1.9, the reader will be led to suspect the following theorem.
166
CHAPTER 4. SYMMETRIC FUNCTIONS
Theorem 4.6.3 Let 4>^ be the character of M>^ evaluated on the class cor-
responding to A. Then
Pa = XI (4-20)
/i>A
Proof. Let A = (Ai, A 2 , . • . , A/). Then we can write equation (4.6) as
+ 0:2' + •••) = X
i fi
Pick out the coefficient of on both sides, where /i = (//i, // 2 , • • • , A^m)- On
the right, it is c\^. On the left, it is the number of ways to distribute the
parts of A into subpartitions A^, . . . , A"’ such that
(4JA* = A and X' \~ fii for alH, (4.21)
i
where equal parts of A are distinguished in order to be considered different in
the disjoint union.
Now consider = 0^(7 t), where tt G <Sn is an element of cycle type A. By
definition, this character value is the number of fixed-points of the action of
7T on all standard tabloids t of shape //. But t is fixed if and only if each cycle
of 7T lies in a single row of t. Thus we must distribute the cycles of length
\i among the rows of length fij subject to exactly the same restrictions as
in (4.21). It follows that c\^ = (/>^, as desired. ■
Equation (4.20) shows that p\ is the generating function for the character
values on a fixed conjugacy class K\ as the module in question varies over all
M^. What we would like is a generating function for the fixed character
of the irreducible Specht module as varies over all conjugacy classes.
To isolate in (4.20), we use the inner product
(0, X) = A y] 0(7r)x(7r) = the multiplicity of x in 0
n! ^
as long as x is irreducible (Corollary 1.9.4, part 2).
If 7T G 5n has type A, then it will be convenient to define the corresponding
power sum symmetric function = P\] similar definitions can be made for
the other bases. For example, if tt = (1, 3, 4)(2, 5), then = P( 3 , 2 )- From
the previous theorem,
Ptt =
To introduce the inner product, we multiply by x^(^)/^* ^nd sum to get
= (Young’s rule)
= s\. (Proposition 4.4.3)
We have proved the following theorem of Frobenius.
4.7. THE CHARACTERISTIC MAP
167
Theorem 4.6.4 If X\- n, then
SA = ^ XI X^{'^)P7T- ■ (4-22)
n!
TVeSn
Note that this is a slightly different type of generating function from those
discussed previously, using the power sum basis and averaging over 5n- Such
functions occur in other areas of combinatorics, notably Polya theory, where
they are related to cycle index polynomials of groups [PTW 83, pages 55-85].
There are several other ways to write equation (4.22). Since is a class
function, we can collect terms and obtain
where and is the value of x^ on Alternatively, we can use
formula (1.2) to express 5 a in terms of the size of centralizers:
4.7 The Characteristic Map
Let = R(Sn) be the space of class functions on Sn- Then there is an
intimate connection between R^ and that we will explore in this section.
First of all, dimR^ = dim A’^ = p(n) (the number of partitions of n), so
these two are isomorphic as vector spaces. We also have an inner product
on R^ for which the irreducible characters on Sn form an orthonormal basis
(Theorem 1.9.3). Motivated by equation (4.22), define an inner product on
A^ by
(«Sa ? ^fj.) ~ ^A/x
and sesquilinear extension (linear in the first variable and conjugate linear in
the second).
We now define a map to preserve these inner products.
Definition 4.7.1 The characteristic map is ch’^ : R^ defined by
ch"(x) = X
/ihn
where X/x is the value of x on the class /x.
It is easy to verify that ch’^ is linear. Furthermore, if we apply ch’^ to the
irreducible characters, then by equation (4.23)
ch"(x^) = s,.
Since ch’^ takes one orthonormal basis to another, we immediately have the
following.
168
CHAPTER 4. SYMMETRIC FUNCTIONS
Proposition 4.7.2 The map ch’^ is an isometry between R^^ and A” . ■
Now consider R = which is isomorphic to A = SnA’^ via the
characteristic map ch = 0nch’^. But A also has the structure of a graded
algebra — i.e., a ring product satisfying (4.5). How can we construct a corre-
sponding product in R^? If X and 'ip are characters of Sn and <Sm, respectively,
we want to produce a character of Sn+m- But the tensor product x^'0 gives
us a character of Sn x Sm, and induction gets us into the group we want.
Therefore, define a product on R by bilinearly extending
X • '0 = (x ® ,
where x and -0 are characters.
Before proving that this product agrees with the one in A, we must gen-
eralize some of the concepts we have previously introduced. Let G be any
group and let A be any algebra over C. Consider functions • G A
with the bilinear form
(x, V')' = 7^ xCffWff ^)-
1^1 g€<3
Note that since g and are in the same conjugacy class in Sn, we have, for
any class function x,
^ m = {x,pY,
ni
neSn
where p : Sn ^ is the function p{7r) = p^r-
Suppose H < G. If X • G -> .4, then define the restriction of x to H to
be the map xin'- H A such that
X\rH (h) = x{h)
for all h E H. On the other hand, if -0 : // — > 4., then let the induction of \p
to G be G -> A defined by
(5) =
I ' xeG
where -0 = 0 outside of H. The reader should verify that the following
generalization of Frobenius reciprocity (Theorem 1.12.6) holds in this setting.
Theorem 4.7.3 Consider H < G with functions x • G A and ip : H ^ A.
If X is a class function on G, then
(V't^-x)' = (^,x4-//)'- ■
This is the tool we need to get at our main theorem about the character-
istic map.
4.8. KNUTH’S ALGORITHM
169
Theorem 4.7.4 The map ch: R-^ A is an isomorphism of algebras.
Proof. By Proposition 4.7.2, it suffices to check that products are preserved.
If X ^nd -0 are characters in Sn and respectively, then using part 2 of
Theorem 1.11.2 yields
ch(x • V’) =
{x-'^,pY
ix<»'t,Pis„xsJ'
TraeSn xSm
cr&Sm
^ E x(^)p-
TTGSn
ch(x) ch(t/;). ■
ml
CrGSrr
4.8 Knuth’s Algorithm
Schensted [Sch 61] realized that his algorithm could be generalized to the
case where the first output tableau is semistandard. Knuth [Knu 70] took
this one step further, showing how to get a Robinson- Schensted map when
both tableaux allow repetitions and thus making connection with the Cauchy
identity [Lit 50, p. 103]. Many of the properties of the original algorithm are
preserved. In addition, one obtains a new procedure that is dual to the first.
Just as we are to allow repetitions in our tableaux, we must also permit
them in our permutations.
Definition 4.8.1 A generalized permutation is a two-line array of positive
integers
i\ ^2 • • •
7T = . .
Jl 32 ••• Jn
whose columns are in lexicographic order, with the top entry taking prece-
dence. Let 7T and ^ stand for the top and bottom rows of tt, respectively. The
set of all generalized permutations is denoted by GP. ■
Note that the lexicographic condition can be restated: ^ is weakly increas-
ing and in tt, jk < jfc+i whenever ik = U+i- An example of a generalized
permutation is
_ 1 1 1 2 2 3
^“ 2 3 3 1 2 1’
(4.24)
If T is any tableau, then let cont T be the content of T. If tt G GP, then tt
and n can be viewed as (one-rowed) tableaux, so their contents are defined.
In the preceding example cont it = (3, 2, 1) and cont tt = (2, 2, 2).
The Robinson-Schensted-Knuth correspondence is as follows.
170
CHAPTER 4. SYMMETRIC EUNCTIONS
Theorem 4.8.2 ([Knu 70]) There is a bijection between generalized permu-
tations and pairs of semistandard tableaux of the same shape,
7T
R-S-K
(T,H)
such that cont re = cont T and cont tt = cont U.
g
Proof, “tt — > (T, [/)” We form, as before, a sequence of tableau pairs
(To,f/o) = ( 0 , 0 ), (Ti,[/i), (T2,t/2), {Tn,Un) = {T,U),
where the elements of n are inserted into the T’s and the elements of tt are
placed in the f/’s. The rules of insertion and placement are exactly the same
as for tableaux without repetitions. Applying this algorithm to the preceding
permutation, we obtain
0 , 2 , 2 3 , 2 3 3 ,
Ti:
133 , 123 , 113
2 2 3 2 2 =T,
3
Ui:
0 ,
11 , 111 , 111 , 111 ,
2 2 2
1 1 1
2 2 =U.
3
It is easy to verify that the insertion rules ensure semistandardness of
T. Also, U has weakly increasing rows because ^ is weakly increasing. To
show that f/’s columns strictly increase, we must make sure that no two equal
elements of tt can end up in the same column. But if = U+i = ^ in the
upper row, then we must have jk < jfc+i- This implies that the insertion path
for jk-\-i will always lie strictly to the right of the path for jk, which implies
the desired result. Note that we have shown that all elements equal to i are
placed in U from left to right as the algorithm proceeds.
g pp
“(T, U) — > 7t” Proceed as in the standard case. One problem is decid-
ing which of the maximum elements of U corresponds to the last insertion.
But from the observation just made, the rightmost of these maxima is the
correct choice with which to start the deletion process.
We also need to verify that the elements removed from T corresponding
to equal elements in U come out in weakly decreasing order. This is an easy
exercise left to the reader. ■
Knuth’s original formulation of this algorithm had a slightly different point
of departure. Just as one can consider a permutation as a matrix of zeros and
ones, generalized permutations can be viewed as matrices with nonnegative
integral entries. Specifically, with each tt G GP, we can associate a matrix of
M = M(7 t) with (z, j) entry
Mij = the number of times Q. j occurs as a column of tt.
4.8. KNUTH’S ALGORITHM
171
Our example permutation has
0 1 2 \
110.
10 0 /
We can clearly reverse this process, so we have a bijection
between GP and Mat == Mat(N), the set of all matrices with nonnegative
integral entries and no final rows or columns of zeros (which contribute noth-
ing to 7t). Note that in translating from tt to M, the number of z’s in tt
(respectively, the number of j’s in tt) becomes the sum of row i (respectively,
column j) of M. Thus Theorem 4.8.2 can be restated.
Corollary 4.8.3 There is a bijection between matrices M G Mat and pairs
of semistandard tableaux of the same shape,
{T,U),
such that cont T and cont U give the vectors of column and row sums, respec-
tively, of M. m
To translate these results into generating functions, we use two sets of vari-
ables, X = {xi, X 2 , . . .} and y = { 2 / 1 , 2/25 • • •}• Let the weight of a generalized
permutation tt be
WtTT = X^y^ = ^cont^ycont^^
To illustrate, the permutation in (4.24) has
i. 3 2 222
WtTT = X 1 X 2 X 32 / 1 2 / 22 / 3 -
If the column Q. j occurs k times in tt, then it gives a contribution of
to the weight. Thus the generating function for generalized permutations is
i:wt,= nE*?»?=n
ttGGP k>Q
1
1 - Xiyj '
As for tableau pairs (T,U), let
Wt(T, U) = X^y^ - ^contC/yContT
as in equation (4.8). Restricting to pairs of the same shape,
shT=shU X \shU=X / \shT=A / A
Since the Robinson-Schensted-Knuth map is weight preserving, we have given
a proof of Cauchy’s formula.
172
CHAPTER 4. SYMMETRIC FUNCTIONS
Theorem 4.8.4 ([Lit 50]) We have
A ij>l
Note that just as the Robinson-Schensted correspondence is gotten by re-
stricting Knuth’s generalization to the case where all entries are distinct, we
can obtain
n! =
Ahn
by taking the coefficient of • • • XnVi • • • 2/n on both sides of Theorem 4.8.4.
Because the semistandard condition does not treat rows and columns uni-
formly, there is a second algorithm related to the one just given. It is called
the dual map. (This is a different notion of “dual” from the one introduced
in Chapter 3, e.g.. Definition 3.6.8.) Let GP' denote all those permutations
in GP where no column is repeated. These correspond to the 0-1 matrices in
Mat.
Theorem 4.8.5 ([Knu 70]) There is a bijection between tt G GP^ and pairs
(T, U) of tableaux of the same shape with T, semistandard,
TT
R-S-K'
(T,U),
such that cont tt = cont T and cont tt = cont U.
Proof, “tt uy> We merely replace row insertion in the R-S-K cor-
respondence with a modification of column insertion. This is done by insisting
that at each stage the element entering a column displaces the smallest entry
greater than or equal to it. For example,
/ 1 1 3 \ 113 3
C2 2 3 - 2 2
\3 / 3
Note that this is exactly what is needed to ensure that T will be column-
strict. The fact that U will be row-strict follows because a subsequence of
7C corresponding to equal elements in tt must be strictly increasing (since
ttgGP').
“(r, U) — > tt” The details of the step-by-step reversal and verification
that 7T G GP' are routine. ■
Taking generating functions with the same weights as before yields the
dual Cauchy identity.
Theorem 4.8.6 ([Lit 50]) We have
^SA(x)sv(y) = R (l+XiPj),
A i,j>l
where X' is the conjugate of X. m
4,8. KNUTH’S ALGORITHM
173
Most of the results of Chapter 3 about Robinson- Schensted have general-
izations for the Knuth map. We survey a few of them next.
Taking the inverse of a permutation corresponds to transposing the associ-
ated permutation matrix. So the following strengthening of Schiitzenberger’s
Theorem 3.6.6 should come as no surprise.
Theorem 4.8.7 If M e Mat and M (T, U), then
Mt R^K ^
We can also deal with the reversal of a generalized permutation. Row and
modified column insertion commute as in Proposition 3.2.2. So we obtain the
following analogue of Theorem 3.2.3.
Theorem 4.8.8 //tt G GP; then T{'K^) = where T' denotes modified
column insertion, m
The Knuth relations become
replace xzy by zxy if x < y < z
and
replace yxz by yzx if x < y < z.
Theorem 3.4.3 remains true.
Theorem 4.8.9 ([Knu 70]) A pair of generalized permutations are Knuth
equivalent if and only if they have the same T -tableau, m
Putting together the last two results, we can prove a stronger version of
Greene’s theorem.
Theorem 4.8.10 ([Gre 74]) Given tt G GP, let shT(7r) = (Ai, A 2 , . . . , A/)
with conjugate (A' 2 , A 2 , . . . , A^). Then for any k, Ai + A 2 -f • • • + A^ and
A'l + A 2 + • • • + A'^ give the lengths of the longest weakly k -increasing and
strictly k-decreasing subsequences of 'k, respectively, m
For the jeu de taquin, we need to break ties when the two elements of
T adjacent to the cell to be filled are equal. The correct choice is forced on
us by semistandardness. In this case, both the forward and backward slides
always move the element that changes rows rather than the one that would
change columns. The fundamental results of Schutzenberger continue to hold
(see Theorems 3.7.7 and 3.7.8).
Theorem 4.8.11 ([Scii 76]) Let T and U be skew semistandard tableaux.
Then T and U have Knuth equivalent row words if and only if they are con-
nected by a sequence of slides. Furthermore, any such sequence bringing them
to normal shape results in the first output tableau of the Robins on- Schensted-
Knuth correspondence. ■
174
CHAPTER 4. SYMMETRIC FUNCTIONS
Finally, we can define dual equivalence, in exactly the same way as
before (Definition 3.8.2). The result concerning this relation, analogous to
Proposition 3.8.1 and Theorem 3.8.8, needed for the Littlewood- Richardson
rule is the following.
Theorem 4.8.12 If T and U are semistandard of the same normal shape,
then T = U. m
4.9 The Littlewood-Richardson Rule
The Littlewood-Richardson rule gives a combinatorial interpretation to the
coefficients of the product s^Si^ when expanded in terms of the Schur basis.
This can be viewed as a generalization of Young’s rule, as follows.
We know (Theorem 2.11.2) that
(4.25)
A
where K\^ is the number of semistandard tableaux of shape A and content
fjL. We can look at this formula from two other perspectives: in terms of
characters or symmetric functions.
If /i h n, then M^ is a module for the induced character But
from the definitions of the trivial character and the tensor product, we have
® ® ® ,
where /i = (/.^i,/i 2 , ■ • • ,/^m)« Using the product in the class function algebra
R (and the transitivity of induction. Exercise 18 of Chapter 1), we can rewrite
(4.25) as
15^1 • 15^2 • • • = Y
A
To bring in symmetric functions, apply the characteristic map to the pre-
vious equation (remember that the trivial representation corresponds to an
irreducible whose diagram has only one row):
S(mi)«(M2) • • • = Y
A
For example,
;^(3,2) ^ 5 ,( 3 , 2) ^(4,1)
with the relevant tableaux being
111 1112 11122
^ • 2 2 ’ 2
This can be rewritten as
1^3 •
^( 3 , 2 ) ^( 4 , 1 ) ^( 5 )
4.9. THE LITTLEWOOD-RICHARDSON RULE
175
or
= -^( 3 , 2 ) + ^( 4 , 1 ) + ^( 5 )-
What happens if we try to compute the expansion
SpSy = (4.26)
A
where /i and u are arbitrary partitions? Equivalently, we are asking for the
multiplicities of the irreducibles in
• x"" = X!
A
or
A
where |/i| + |i/| = n. The are called the Littlewood- Richards on coefficients.
The importance of the Littlewood-Richardson rule that follows is that it gives
a way to interpret these coefficients combinatorially, just as Young’s rule does
for one- rowed partitions.
We need to explore one other place where these coefficients arise: in the
expansion of skew Schur functions. Obviously, the definition of s\{x) given
in Section 4.4 makes sense if A is replaced by a skew diagram. Furthermore,
the resulting function is still symmetric by the same reasoning as
in Proposition 4.4.2. We can derive an implicit formula for these new Schur
functions in terms of the old ones by introducing another set of indeterminates
y = (j/1,2/2,---)-
Proposition 4.9.1 ([Mac 79]) Define s\{x.,y) = sx(xi,X 2 , ■ ■ ■ ,yi,y 2 , ■ ■ ■)■
Then
sa ( x , y) = Yl s^,{x.)sx/^,{y). (4.27)
fiCX
Proof. The function 5 a (x, y) enumerates semistandard fillings of the diagram
A with letters from the totally ordered alphabet
{K2<3<---<T<2'<3' <•••}.
In any such tableau, the unprimed numbers (which are weighted by the x’s)
form a subtableau of shape in the upper left corner of A, whereas the
primed numbers (weighted by the y’s) fill the remaining squares of A//i. The
right-hand side of (4.27) is the generating function for this description of the
relevant tableaux. ■
Since is symmetric, we can express it as a linear combination of
ordinary Schur functions. Some familiar coefficients will then appear.
Theorem 4.9.2 If the are Littlewood-Richardson coefficients, where
\iy\ = |A|, then
176
CHAPTER 4. SYMMETRIC FUNCTIONS
Proof. Bring in yet a third set of variables z = {zi, Z 2 , ■ ■ ■} ■ By using the
previous proposition and Cauchy’s formula (Theorem 4.8.4),
^S^(x)SA/^(y)SA(z) =
Taking the coefficient of s^(x)Si,(y)5A(z) on both sides and comparing with
equation (4.26) completes the proof. ■
One last definition is needed to explain what the count.
Definition 4.9.3 A ballot sequence^ or lattice permutation^ is a sequence of
positive integers tt = ^122 • • • such that, for any prefix TTk = i\i 2 • • - ik and
any positive integer /, the number of /’s in tt/c is at least as large as the number
of (/ -f l)’s in that prefix. A reverse ballot sequence or lattice permutation is
a sequence tt such that tt’’ is a ballot sequence. ■
As an example,
TT = 1 1 2 3 2 1 3
is a lattice permutation, whereas
TT = 1 2 3 2 1 1 3
is not because the prefix 1 2 3 2 has more twos than ones.
The name ballot sequence comes from the following scenario. Suppose
the ballots from an election are being counted sequentially. Then a ballot
sequence corresponds to a counting where candidate one always (weakly)
leads candidate two, candidate two always (weakly) leads candidate three,
and so on.
Furthermore, lattice permutations are just another way of encoding stan-
dard tableaux. Given P standard with n elements, form the sequence tt =
21^2 ... in? where ik =i\ik appears in row i of P. The fact that the entries less
than or equal to k form a partition with weakly decreasing parts translates to
the ballot condition on tt^. It is easy to construct the inverse correspondence
and see that our example lattice permutation codes the tableau
1 2 6
P- 3 5
4 7
We are finally ready to state and prove the Littlewood-Richardson rule.
Although it was first stated by these two authors [L-R 34], complete proofs
^SA(x,y)SA(z)
A
n — — -
I — XiZj I — I
ViZj
^s^(x)s^(z)
Y^s^{y)su{z)
Y^s^{K)s^{y)Sf,{z)s^{z).
4.9. THE LITTLEWOOD-RICHARDSON RULE
177
were not published until comparatively recently by Thomas [Tho 74, Tho 78]
and Schiitzenberger [Scii 76]; our demonstration is based on the latter.
Theorem 4.9.4 (Littlewood-Richardson Rule [L-R 34]) The value of
the coefficient is equal to the number of semistandard tableaux T such
that
1. T has shape X/fi and content v,
2. the row word ofT, ttt, is a reverse lattice permutation.
Proof. Let be the number of tableaux T satisfying the restrictions of the
theorem. Then we claim that it suffices to find a weight-preserving map
T (4.28)
from semistandard tableaux T of shape A/// to semistandard tableaux U of
normal shape such that
1. the number of tableaux T mapping to a given U of shape u depends
only on A, fi, and v and not on U itself;
2. the number of tableaux T mapping to a particular choice of U is
If such a bijection exists, then
shT=A//i shU=v
Comparison of this expression with Theorem 4.9.2 and the fact that the Schur
functions are a basis completes the proof.
It turns out that the map j needed for equation (4.28) is just the jeu de
taquin! To show that j satisfies property 1, consider U and U' of the same
shape V. We will define a bijection between the set of tableaux T that map
to U and the set of T' that map to U' as follows. Let P be a fixed standard
tableau of shape p. Then, in the notation of Section 3.8, we always have
U = ffi{T). Let the vacating tableau for this sequence of slides be denoted
by Qt? a standard tableau of shape \/u. Consider the composition
T — >U' (4.29)
i.e., slide T to normal shape to obtain U and Qt, replace {7 by I/', and then
slide U' back out again according to the vacating tableau.
We first verify that this function is well defined in that T' must be sent to
U' by jeu de taquin and must have shape X/p. The first statement is clear,
since all the slides in jQrj, are reversible. For the second assertion, note that
by the definition of the vacating tableau,
T = 3Qrf{T)=jQ^{U).
178
CHAPTER 4. SYMMETRIC EUNCTIONS
Also,
T'=3q,{U').
Since U and f/' are of the same normal shape, they are dual equivalent (The-
orem 4.8.12), which implies that
T ^ T'. (4.30)
In particular, T and T' must have the same shape, as desired.
To show that (4.29) is a bijection, we claim that
is its inverse, where Qt> = v^{T'). But equation (4.30) implies that the
corresponding vacating tableaux Qt and Qt' are equal. So
fiT')=j^jQAU')=j^3Qr>{U') = U'
and
3Qr, (U) = 3Qr-f{T) = 3Qrf{T) = T,
showing that our map is a well-defined inverse.
Now we must choose a particular tableau Uq for property 2. Let Uo be
the superstandard tableau of shape and content z/, where the elements of
the ith row are all i’s. Notice that the row word ttuq is a reverse lattice
permutation. In considering all tableaux mapped to Uq by j, it is useful to
have the following result.
Lemma 4.9.5 If we can go from T to T' by a sequence of slides, then ttt is
a reverse lattice permutation if and only if ttt' is.
Proof. It suffices to consider the case where T and T' differ by a single move.
If the move is horizontal, the row word doesn’t change, so consider a vertical
move. Suppose
Ri
X
Rr
Si
•
Sr
and
Ri
•
Rr
Si
X
Sr
where i?/ and 5/ (respectively, Rr and Sr) are the left (respectively, right)
portions of the two rows between which x is moving. We show that if cr =
4.10. THE MURNAGHAN-NAKAYAMA RULE
179
is a ballot sequence, then so is cr' == . (The proof of the reverse implication
is similar.)
Clearly, we need only check that the number of (x -f l)’s does not exceed
the number of x’s in any prefix of cr' ending with an element of Ri or Sr- To
do this, we show that each x + 1 in such a prefix can be injectively matched
with an x that comes before it. If the x + 1 occurs in Rr or a higher row, then
it can be matched with an x because (j is a ballot sequence. Notice that all
these x’s must be in higher rows because the x’s in Rr come after the (x-h l)’s
in Rr when listed in the reverse row word. By semistandardness, there are
no (x + l)’s in Ri to be matched. For the same reason, every x + 1 in 5r must
have an x in Rr just above it. Since these x’s have not been previously used
in our matching, we are done. ■
Now note that Uq is the unique semistandard tableau of normal shape u
whose row word is a reverse lattice permutation. Thus by the lemma, the
set of tableaux T with shape X/fi such that j{T) = Uq are precisely those
described by the coefficients This finishes the proof of the Littlewood-
Richardson rule. ■
As an illustration, we can calculate the product 5(2,i)5(2,2)« Listing all
tableaux subject to the ballot sequence condition with content (2,2) and
skew shape A/ (2, 1) for some A yields
• • 11 , • • 1 1 , • • 1 , • • 1 , • • 1 , • •.
• 22 m2 • 1 2 • 1 •2 •!
2 2 2 2 1 1 2
2 2
Thus
5 ( 2 , 1 ) 5 ( 2 , 2 ) = '^( 4 , 3 ) + 5 ( 4 ^ 2 , 1 ) + 5 ( 32 ^ 1 ) + ^{ 3 , 2 ^) + 5 ( 3 , 2 , 12 ) + 5 ( 23 , 1 ).
For another example, let us find the coefficient of 5 ( 5 , 3 , 2 , 1 ) in 5 ( 3 , 2 , i)S( 3 , 2 )-
First we fix the outer shape A = (5, 3, 2, 1) and then find the tableaux as
before:
• •• 11 , •••!!,
• • 1 • • 2
• 2 • 1
2 2
• •• 11 .
• • 2
• 2
1
It follows that
( 5 , 3 , 2 , 1 )
^( 3 , 2 , 1 )( 3 , 2 )
3.
4.10 The Murnaghan-Nakayama Rule
The Murnaghan-Nakayama rule [Mur 37, Nak 40] is a combinatorial way of
computing the value of the irreducible character on the conjugacy class a.
The crucial objects that come into play are the skew hooks.
180
CHAPTER 4. SYMMETRIC FUNCTIONS
Definition 4.10.1 A skew hook^ or rim hook, is a skew diagram obtained
by taking all cells on a finite lattice path with steps one unit northward or
eastward. Equivalently, ^ is a skew hook if it is edgewise connected and
contains no 2 x 2 subset of cells:
(4.31)
Generalizing the definition in Section 3.10, the leg length of ^ is
ll{^) = (the number of rows of — 1. ■
For example.
has ll{^) = 4. The name rim hook comes from the fact that such a diagram can
be obtained by projecting a regular hook along diagonals onto the boundary
of a shape. Our example hook is the projection of H\^i onto the rim of
A = (6,3, 3, 1,1).
project
Notice that we are using single Greek letters such as ^ for rim hooks even
though they are skew. If ^ = A/^u, then we write A\^ for fx. In the preceding
example, fx = (2, 2). Also, if a = (ai, Q 2 , . . . , a^) is a composition, then let
a\ai — (a 2 , . . . , a^)- With this notation, we can state the main result of this
section.
Theorem 4.10.2 (Murnaghan-Nakayama Rule [Mur 37, Nak 40]) If
X is a partition ofn and a = (oi, . . . , a^) 'Is a composition ofn, then we have
where the sum runs over all rim hooks ^ of X having a\ cells.
Before proving this theorem, a few remarks are in order. To calculate Xa
the rule must be used iteratively. First remove a rim hook from A with a\
cells in all possible ways such that what is left is a normal shape. Then strip
away hooks with Q 2 squares from the resulting diagrams, and so on. At some
stage either it will be impossible to remove a rim hook of the right size (so
4.10. THE MURNAGHAN-NAKAYAMA RULE
181
the contribution of the corresponding character is zero), or all cells will be
deleted (giving a contribution of ±X(o) =
To illustrate the process, we compute X(s 4 ’ 2 )- The stripping of hooks can
be viewed as a tree. Cells to be removed are marked with a dot, and the
appropriate sign appears to the right of the diagram:
A =
v/
\
+1
0
(-1)3
The corresponding calculations are
^(4,4.3)
^(5,4,2)
_y(4.2) , (3,2,1)
A.(4,2) ^ A.(4,2)
-(-X(2f^) +0
-(-(-xg!))
(-1)3.
Also note that the branching rule from Section 2.8 is a special case of
Murnaghan-Nakayama. Take a = (1, a 2 , . . . , a^) and let tt E Sn have type a.
Since tt has a fixed-point,
Xa = X^(7 t) = X''4-S„_i (tt),
which corresponds to the left-hand side of the first equation in Theorem 2.8.3.
As for the right side, |^| = 1 forces A\^ to be of the form A“ with all signs
(-1)0 = +!.
182
CHAPTER 4. SYMMETRIC FUNCTIONS
We now embark on a proof of the Murnaghan-Nakayama rule. It is based
on the one in James [Jam 78, pp. 79-83].
Proof (of Theorem 4.10.2). Let m = ai. Consider ttct G Sn-m S <5^,
where tt has type ( 02 ? • • • ? and a is an m-cycle. By part 2 of Theo-
rem 1.11.3, the characters where /j,\- n — m, z/ h m, form a basis for
the class functions on Sn-m x <5^- So
Xa = X^(Tro-) = (ttct) = (4.32)
/ihn — m
i/hm
To find the multiplicities, we use Probenius reciprocity (Theorem 1.12.6)
and the characteristic map:
= (X^i5„_„xs„,x''«'x‘'>
= (x\(x"®x")t^")
= ix^,x^-x'')
=
where is a Littlewood-Richardson coeflficient. Thus we can write equation
(4.32) as
X^(7ra-) = x^i^) Y ^^yX‘'(<^)- (4-33)
^hn—m yhm
Now we must evaluate where cr is an m-cycle.
Lemma 4.10.3 If v\- m, then
(0 otherwise.
Note that this is a special case of the theorem we are proving. If a = (m),
then Xa 7^ 0 when we can remove all the cells of z/ in a single sweep — that
is, when u itself is a hook diagram (r, 1’^“’^). In this case, the Murnaghan-
Nakayama sum has a single term with leg length m — r.
Proof (of Lemma 4.10.3). By equation (4.23), x’lm) ^{m) = ^ times the
coefficient of Pm in
Using the complete homogeneous Jacobi-Trudi determinant (Theorem 4.5.1),
we obtain
Sjy = \hi/^—i^j\ixi — ^ ^ dih/t,
K
where the sum is over all compositions k = (z^i, . . . , /^/) that occur as a term
in the determinant. But each h^. in h^ can be written as a linear combination
of power sums. So, since the p’s are a multiplicative basis, the resulting linear
4.10. THE MURNAGHAN-NAKAYAMA RULE
183
combination for will not contain Pm unless k contains exactly one nonzero
part, which must, of course, be m. Hence X(m) ^ ^ when hm appears-in
the preceding determinant.
The largest index to appear in this determinant is at the end of the first
row, and - 1 -f / = the hooklength of cell (1, 1). Furthermore, we
always have m = \u\ > hi^i. Thus X(m) nonzero only when hi^i = m, i.e.,
when z/ is a hook (r, In this case, we have
hm
hi ...
ho hi
— (—1)’^ ^hm+ other terms not involving Pm-
But hm = S(^m) corresponds to the trivial character, so comparing coefficients
of Pml'm in this last set of equalities yields X(m) ~ desired. ■
In view of this lemma and equation (4.33), we need to find when i' is
a hook. This turns out to just be a binomial coefficient.
Lemma 4.10.4 Let X h n, p. \~ n — m, and u = (r, Then = 0
unless each edgewise connected component of X/p is a rim hook. In that case,
if there are k component hooks spanning a total of c columns, then
Proof. By the Littlewood- Richardson rule (Theorem 4.9.4), is the number
of semistandard tableaux T of shape X/p containing r ones and a single copy
each of 2, 3, . . . , m - r + 1 such that ttt is a reverse lattice permutation. Thus
the numbers greater than one in must occur in increasing order. This
condition, together with semistandardness, puts the following constraints on
T:
Tl. Any cell of T having a cell to its right must contain a one.
T2. Any cell of T having a cell above must contain an element bigger than
one.
If T contains a 2 x 2 block of squares, as in (4.31), then there is no way
to fill the lower left cell and satisfy both Tl and T2. Thus c^j^ = 0 if the
components of the shape of T are not rim hooks.
Now suppose X/p = |+)JL^ where each is a component skew hook.
Conditions Tl, T2 and the fact that 2 through m - r + 1 increase in show
hr . • •
ho hi
0 ho
0 0
184
CHAPTER 4. SYMMETRIC FUNCTIONS
that every rim hook must have the form
(i)
1
1
1
b
d
1
1
d + 1
d+2
d + 3
where d > 1 is the smallest number that has not yet appeared in and b is
either 1 or d — 1. Thus all the entries in are determined once we choose
the value of b. Furthermore, in we must have b — 1. By T2 we have
c — (A: — 1) ones already fixed in T. Hence there are r — c + A; — 1 ones left to
distribute among the A: - 1 cells marked with a b. The number of ways this
can be done is
=( up-'
\r-c + k-lj \c-r
Putting the values from the two lemmas in equation (4.33) we obtain
fji r=l ^ ^
Now, k < c < m, since these three quantities represent the number of skew
hooks the number of columns in the and the number of cells in the
respectively. Thus, using Exercise Id of this chapter.
k-1
0
r)
+ ■
k-1
k-1
0
if A: - 1 = 0,
otherwise.
But if A: = 1, then A//i is a single skew hook ^ with m squares and c
columns. Hence m — c = so equation (4.34) becomes
X^{na) = ^
|4l=m
finishing the proof of the Murnaghan-Nakayama rule. ■
It is possible to combine the calculations for Xa individual tableaux.
Definition 4.10.5 A rim hook tableau is a generalized tableau T with posi-
tive integral entries such that
1. rows and columns of T weakly increase, and
2. all occurrences of i in T lie in a single rim hook. ■
4.11. EXERCISES
185
Obviously, the stripping of rim hooks from a diagram determines a rim
hook tableau by labeling the last hook removed with ones, the next to last
with twos, and so on. For example, the tableau corresponding to the left
branch of the preceding tree is
12 2 2
T= 1233 .
3 3 3
Now define the sign of a rim hook tableau with rim hooks to be
(-If = n
The tableau just given has sgn(T) = (—1)^- It is easy to see that the
Murnaghan-Nakayama theorem can be restated.
Corollary 4.10.6 Let X be a partition of n and let a = (oi, . . . ^ak) he any
composition of n. Then
Xa =
T
where the sum is over all rim hook tableaux of shape A and content a. ■
It is worth mentioning that Stanton and White [S-W 85] have extended
Schensted’s construction to rim hook tableaux. This, in turn, has permit-
ted White [Whi 83] to give a bijective proof of the orthonormality of the
irreducible characters of Sn (Theorem 1.9.3) analogous to the combinatorial
proof of
n! = Y.T?,
Ahn
which is just the fact that the first column of the character table has unit
length.
4.11 Exercises
1. Recall that the binomial coefficients are defined by
= the number of ways to choose 5 C {1, 2, . . . , n} with \S\ = k.
Note that this definition makes sense for all n > 0 and all integers k
where (^) = 0 if A: < 0 or fc > n.
(a) Show that for n > 1 we have the recursion
n - 1
k
-h
n — 1
k-1
with boundary condition (^) = 5k,o-
186
CHAPTER 4. SYMMETRIC FUNCTIONS
(b) Show that for 0 < /c < n we have
fTi\ _ n\
\kj ^ k!(n-k)l
in two different ways: inductively and by a direct counting argu-
ment.
(c) Show that
in two different ways: inductively and by using weight generating
functions.
(d) Show that
where (Jri,o is the Kronecker delta, in two different ways: using the
previous exercise and inductively.
(e) Let 5 be a set. Show that
TCS
in two different ways: using the previous exercise, and by assigning
a sign to every TCS and then constructing a fixed-point free,
sign-reversing involution on the set of subsets of S.
(f) Prove Vandermonde’s convolution
§C)G-^cr)
in three different ways: inductively, using generating functions,
and by a counting argument.
2. Prove the following theorem of Glaisher. The number of partitions of
n with no part divisible by k is equal to the number of partitions of n
with each part appearing at most k — 1 times. Which theorem in the
text is a special case of this result?
3. Let p(n, k) (respectively, Pd{n, k)) denote the number of partitions of n
into k parts (respectively, k distinct parts). Show that
’Y^p{n,k)x^t’^
n>0
k>0
n
2>1
1
1 — xH
4.11. EXERCISES
187
'^Pd{n,k)x'^t^ = JJ(l + x*i)
n>0 i>l
fe >0 “
;^(l-x)(l-a;2)---(l-x")’
4. Consider the plane V = {{i,j) : j > 1}, i«e., the “shape” consisting
of an infinite number of rows each of infinite length. A plane partition
of n is an array, T, of nonnegative integers in V such that T is weakly
decreasing in rows and columns and Y!,(ij)ev ~ example,
the plane partitions of 3 (omitting zeros) are
3, 2 1, 2, 111, 11, 1.
1 1 1
1
(a) Let
pp{n) = the number of plane partitions of n
and show that
n>0 i>l ^ ’
in two ways: combinatorially and by taking a limit.
(b) Define the trace of the array T to be Yli Let
pp{n^ k) = the number of plane partitions of n with trace k
and show that
n,fc>0 i>l ^ ^
5. (a) Let r be a rooted tree (see Chapter 3, Exercise 18). Show that the
generating function for reverse r-partitions is
TT ^
11 1 _
V^T
and use this to rederive the hook formula for rooted trees.
(b) Let A* be a shifted shape (see Chapter 3, Exercise 21). Show that
the generating function for reverse A*-partitions is
n
(iJ)eX’
1
1 —
and use this to rederive the hook formula for shifted tableaux.
188
CHAPTER 4. SYMMETRIC FUNCTIONS
6. The principal specialization of a symmetric function in the variables
{x\,X 2 , • > • ,Xm} is obtained by replacing Xi by for all i.
(a) Show that the Schur function specialization s\{q,q^ ^ , q^) is the
generating function for semistandard A-tableaux with all entries of
size at most m.
(b) Define the content of cell (i, j) to be
^i,j — J
Prove that
1 —
where m(A) = Yli>i Use this result to rederive Theorem 4.2.2.
7. Prove part 2 of Theorem 4.3.7.
8. Let Ag denote the ring of symmetric functions of degree n with integral
coefficients. An integral basis is a subset of Ag such that every element of
Ag can be uniquely expressed as an integral linear combination of basis
elements. As A varies over all partitions of n, which of the following are
integral bases of Ag; tha, Pa? ^a, hx, s\7
9. (a) Let
efc(n) = ek(xi,X 2 ,...,Xn)
and similarly for the homogeneous symmetric functions. Show that
we have the following recursion relations and boundary conditions.
For n > 1,
efe(n) = efe(n - 1) + a;„efe_i(n - 1),
hk{n) = hk{n - 1) + Xnhk-i{n),
and efc(O) = hk{0) = Sk,o^ where Sk^o is the Kronecker delta.
(b) The (signless) Stirling numbers of the first kind are defined by
c(n, k) = the number of tt G <Sn with k disjoint cycles.
The Stirling numbers of the second kind are defined by
5(n, k) = the number of partitions of {1, . . . , n} into k subsets.
Show that we have the following recursion relations and boundary
conditions. For n > 1,
c(n, k) = c{n - 1, fc — 1) -f (n — l)c(n — 1, k)
S{n,k) = S{n - l^k - 1) kS{n — l,k)
and c(0, k) = S{0, k) = dk,o-
4.11. EXERCISES
189
(c) For the binomial coefficients and Stirling numbers, show that
(fc) "" efc(l,l,...,l)
n— /c+l
c{n,k) = e„_fc(l,2, 1),
S{n,k) = h„-k{l,2,...,k).
10. Prove the elementary symmetric function version of the Jacobi-Trudi
determinant.
11. A sequence {ak)k>o = ao,ai,a 2 , . • . of real numbers is log concave if
afc-iOfc+i < a\ for all A; > 0.
(a) Show that if a/c > 0 for all A:, then log concavity is equivalent to
the condition
Ok-iai^i < Okai for all / > A; > 0.
(b) Show that the following sequences of binomial coefficients and Stir-
ling numbers (see Exercise 9) are log concave in two ways: induc-
tively and using results from this chapter.
(c(n, A:))jfc>o, (S(n, fc))n>o-
(c) Show that if (xn)n>o is a log concave sequence of positive reals
then, for fixed A;, the sequences
(e/c(Xi, ... 7 ^n))n>0 and • • • 5 ^n))n>0
are also log concave. Conclude that the following sequences are
log concave for fixed n and k:
(c{n + j, fc + {S{n +j,k + j))j>o-
12. (a) Any two bases {u\ : A h n} and : A h n} for A’^ define a
unique inner product by
{u\^ Vfj^) =
and sesquilinear extension. Also, these two bases define a generat-
ing function
/(x,y) = Y^ux{K)vx{y).
Ahn
Prove that two pairs of bases define the same inner product if and
only if they define the same generating function.
190
CHAPTER 4. SYMMETRIC FUNCTIONS
(b) Show that
= n
A if
1
1 - XiVj ■
13. Prove the following. Compare with Exercise 7 in Chapter 3.
(a) Theorem 4.8.7
(b) There is a bijection M < — > T between symmetric matrices M e
Mat and semistandard tableaux such that the trace of M is the
number of columns of odd length of T.
(c) We have the following equalities:
A
A' even
.1. 1 7* . ii ii 1 T • T • ^
i i<j
n — ^ —
f 1 — XiXj
14. Give a second proof of the Littlewood-Richardson rule by finding a
bijection
(P,Q)^(T,U)
such that
• P, Q, T, and U are semistandard of shape /i, z/, A, and A//i,
respectively,
• cont P + cont Q = cont T, where compositions add componentwise,
and
• TTt/ is a reverse lattice permutation of type u.
15. The Durfee size of a partition A is the number of diagonal squares (i, i)
in its shape. Show that Xa = 0 if Durfee size of A is bigger than the
number of parts of a.
Chapter 5
Applications and
Gener alizat ions
In this chapter we will give some interesting applications and extensions of
results found in the rest of the book. Before tackling this material, the reader
may find it helpful to review the poset definitions and examples at the begin-
ning of Section 2.2 and Definition 2.5.7.
We begin with a couple of sections studying differential posets, which were
introduced by Stanley [Stn 88]. This permits us to give two more proofs of
the sum-of-squares formula (5.1). One uses linear algebra and the other, due
to Fomin [Fom 86], is bijective. But both of them generalize from the poset
called Young’s lattice (the classical case) to the wider class of differential
posets.
Next comes a pair of sections concerning groups acting on posets [Stn 82].
In the first we study some interesting modules that arise from the symmetric
group acting on the Boolean algebra. The second shows how group actions
can be used to prove unimodality results.
The final section introduces graph colorings. In particular, it shows how
certain symmetric functions of Stanley [Stn 95] can be used to study such
objects.
5.1 Young’s Lattice and Differential Posets
We are going to give a third proof of the formula
= (5.1)
Ahn
based on properties of a poset called Young’s lattice. Two interesting points
about this demonstration are that it uses linear algebra and generalizes to
the larger family of differential partially ordered sets.
191
192
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Definition 5.1.1 Let {A, <) be a partially ordered set. Then a^b e A have
a greatest lower bound, or meet ,if there is c e A such that
1. c < a, c < b and 2. if d < a, d < b, then d < c. (5.2)
The meet of a and 6, if it exists, is denoted by a A 6. The least upper bound,
or join, of a and b is defined by reversing all the inequalities in (5.2) and is
written a V 5. A poset where every pair of elements has a meet and a join is
called a lattice, m
As examples, the Boolean algebra is a lattice, since if 5, T G Bn then
S/\T = SnT and 5VT = 5UT
This relationship between A and fl, as well as V and U, makes the symbols
for meet and join easy to remember. The n-chain Cn is also a lattice, since if
hj ^ Cn then
i A j = min{ 2 , j } and iy j = max{i, j } .
The lattice that concerns us here was first studied by Kreweras [Kre 65].
Definition 5.1.2 Young’s lattice, Y, is the poset of all integer partitions
ordered by inclusion of their Ferrers diagrams. ■
Here is a picture of the bottom portion of Y (since it is infinite, it would be
hard to draw the whole thing!) :
\ \ t \ /
\ Z'
□
t
0
It is easy to verify that F is a lattice with A and V being intersection and
union of Ferrers diagrams. Note that Y has a minimum, namely the empty
partition 0 of 0, but no maximum. Here are a couple of other basic properties
of Young’s lattice.
\
\
5.1. YOUNG^S LATTICE AND DIFFERENTIAL POSETS
193
Proposition 5.1.3 The poset Y satisfies the following two conditions.
(1) If X G Y covers k elements for some k, then it is covered by k 1
elements.
(2) If X ^ and X and /a both cover I elements for some I, then they are
both covered by I elements. In fact, I < 1.
Proof. (1) The elements covered by A are just the partitions A~ obtained by
removing an inner corner of A, while those covering A are the A*^ obtained
by adding an outer corner. But along the rim of A, inner and outer corners
alternate beginning and ending with an outer corner. So the number of A“^
is one greater than the number of A~.
(2) If X,n both cover an element of Y, then this element must be A A /z.
So / < 1. Also, / = 1 if and only if |A| = \fi\ = n and |A fl = n — 1 for some
n. But this is equivalent to |A| = \p\ = n and |A U /x| = n + 1, i.e., A,/i are
both covered by a single element. ■
Note that each of the implications in this proposition is equivalent to its
converse, so they could actually have been stated as “if and only if.” We can
rephrase these two conditions in terms of linear operators.
Definition 5.1.4 Consider the vector space CY of formal linear combina-
tions of elements of Y. Define two linear operators, down, D, and up, U,
by
D{\) = ^ A-,
A--<A
U{\) = A+,
A+^A
and linear extension. ■
For example
Proposition 5.1.5 The operators D,U satisfy
DU -UD = I, ( 5 . 3 )
where I is the identity map.
194
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Proof. By linearity, it suffices to prove that this equation holds when applied
to a single A G CY. But by the previous proposition
DI7(A) = ^^ + (A; + l)A,
where the sum is over all ^ ^ X such that there is an element of Y covering
both, and A; + 1 is the number of elements covering A. In the same way,
UD{^) = Y,^^’ + k\,
and taking the difference completes the proof. ■
Equation (5.3) is a special case of the next result.
Corollary 5.1.6 Let p{x) G C[x] be a polynomial. Then
Dp{U)=p\U)+p{U)D,
where p'{x) is the derivative.
Proof. By linearity, it suffices to prove this identity for the powers n > 0.
The case n = 0 is trivial. So, applying induction yields
= (DU^)U
= + U^D)U
= nU^pU^(IPUD)
= + ■
It is also possible to obtain tableaux from Young’s lattice using chains.
Definition 5.1.7 Let ^ be a poset and consider a < b in A. An a-b (as-
cending) chain of length n is a sequence
C : a = Oq < a\ < a\ < - • ' < On — b. (5.4)
A descending chain is defined analogously by reversing all the inequalities.
An a-b chain is saturated if it is contained in no larger a-b chain. ■
So a chain of length n is just a copy of Cn in A. In Y , a chain is saturated
if and only if each of the inequalities in (5.4) is a cover. Also, there is a
simple bijection between standard A-tableaux P and saturated 0-A chains
0 = A^ ^ A^ -<•••-< A’^ = A given by
A* = shape of the tableau of elements < i in P.
(This is essentially the same as the way we converted a tabloid into a sequence
of compositions in Section 2.5.) For example, the tableau
1 3
P= 2 5
4
5.1. YOUNG^S LATTICE AND DIFFERENTIAL POSETS
195
corresponds to the saturated chain
• • • • •
• •
One could just as well put saturated descending chains in bijection with
tableaux, or in terms of the D and U operators and partitions A h n,
£>«(A) =
i7"(0) =
Ahn
Also, a pair of tableaux of the same shape A can be viewed as a saturated
0-A chain followed by one from A back to 0. So
D"C/“(0) = 0^(/^)^ (5.5)
Ahn
We now have all the ingredients to give another proof of the sum-of-squares
formula.
Theorem 5.1.8 We have
Ahn
Proof. In view of equation (5.5) we need only show that D'^U'^{0) = n!0. We
will do this by induction, it being clear for n = 0. But using Corollary 5.1.6
and the fact that 0 is the minimal element of Y, we obtain
D^U^{0) = D^-\DU^){0)
= D^-\nU^-^ -hU^D){9)
= nD^-^U^-\0) + D^-'^U^D{0)
= n{n - 1)!0 -f 0. ■
It turns out that all these results for Young’s lattice can be generalized to
a wider family of posets, as has been done by Stanley [Stn 88, Stn 90]. (See
also Fomin [Fom 94].) First, however, we need a few more definitions.
Definition 5.1.9 Let A be a poset having a minimum element, 0. Then A is
graded if, for each a e A, all 6-a saturated chains have the same length. We
denote this length by rka, called the rank of a. Furthermore, the kth rank of
A is the set
Ak = {a e A : rka = k}.
Finally, we say that A has rank n if there is a nonnegative integer n such that
n = max{rka : a G A}. ■
196
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
The posets Cn, Bm and V are all graded. Their A:th ranks are the number k,
all subsets of {1, . . . ,n} with k elements, and all partitions of k, respectively.
The first two posets have rank n, but Y has arbitrarily long chains and so
has no rank.
Definition 5.1.10 ([Stn 88]) A poset A is differential if it satisfies the fol-
lowing three conditions.
DPI A has a (unique) minimum element 0 and is graded.
DP2 U a e A covers k elements for some A:, then it is covered by fc + 1
elements.
DPS If a ^ 6 and a and b both cover I elements for some /, then they are
both covered by I elements. ■
Actually, these posets are called 1-diflPerential by Stanley. For the definition
of his more general r-differential posets, see Exercise 6. By Proposition 5.1.3,
y is a differential poset. For another example, see Exercise 2.
As with y, we can show that the integer I in DPS is at most 1.
Lemma 5.1.11 If A is a poset satisfying DPI and DPS, then I <1.
Proof. Suppose to the contrary that there are elements a,b E A with / > 2,
and pick a pair of minimal rank. Then a, b both cover c, d one rank lower.
But then c,d are both covered by a,b and so have an I value at least 2, a
contradiction. ■
We wish to define up and down operators in any differential poset A, But
first we need a lemma to make them well defined.
Lemma 5.1.12 If A is differential, then the nth rank. An, is finite for all
n > 0.
Proof. Induct on n, where the case n = 0 is taken care of by DPI. Assume
that all ranks up through An are finite. Any a E An can cover at most |A„_i|
elements, and so can be covered by at most |An-i| + 1 elements by DP2. It
follows that
|An+l| < \An\ (|An-l| + 1)
forcing An-^-i to be finite. ■
Definition 5.1.13 Let A be a poset where, for any a £ A, the sets {be
A : b ^ a} and {b e A : b >- a} are finite. Consider the vector space CA
of formal linear combinations of elements of A. Define two linear operators,
down, D, and up, U, in CA by
D{a) =
b-<a
U{bl) =
b>-a
5.2. GROWTHS AND LOCAL RULES
197
and linear extension. Note that if A is differential, then by Lemma 5.1.12
these operators are well defined. ■
The proof of the next result is similar to that of Proposition 5.1.5 and so
is left to the reader.
Proposition 5.1.14 Let A he a graded poset with A^ finite for all n > 0.
Then the following are equivalent.
1. A is differential.
2. DU-UD = I.m
In fact, differential posets were so named because of the identity DU — UD =
I. Specifically, if we let D be differentiation and x be the multiplication by x
operator acting on the polynomial ring C[x], then Dx — xD = I. The algebra
of operators generated by D and x modulo this relation is called the Weyl
algebra.
We can now generalize Theorem 5.1.8. Again, the demonstration is close
to that of the original result, and so the reader will have no trouble filling in
the details.
Theorem 5.1.15 ([Stn 88]) In any differential poset A,
E
a^An
where is the number of saturated 0-a chains. ■
5.2 Growths and Local Rules
It would be nice to give a bijective proof of Theorem 5.1.15 as the Robinson-
Schensted algorithm does for Theorem 5.1.8. This was done as part of a
theory developed by Fomin [Fom 86, Fom 95], which we now present. For
other expositions, see Roby’s thesis [Rob 91] or Stanley’s text [Stn 99, pp.
327-330].
The basic definition is as follows.
Definition 5.2.1 ([Fom 95]) Let be posets. Then a growth is a func-
tion g : A ^ B that preserves the relation “covered by or equal to,” which
means
a:<b=> g{a) ^ g{b). ■
Note that such a function is order preserving, i.e., a < b implies g{a) < g{b),
but need not be injective. The term “growth” comes from biology: If A = Cn,
then we can think of A’s elements as units of time and the value g{i) as the
state of the system studied at time i. So at time z -(- 1 the system either stays
the same or grows to a subsequent level.
198
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Given posets A, define their product^ A x B, as having elements (a, b)
with a e A, b e B ordered by
(a, b) < (a', b') in A X B if a < a' in A and b < b' in B.
We will be particularly concerned with growth functions g : C^^ A where
C^ = Cn X Cn and A is differential. We will represent the elements of C^ as
vertices (^, j), 0 < i, j < n, in the first quadrant of the plane and the covering
relations as lines (rather than arrows) connecting the vertices. The squares
thus formed will also be coordinatized, with square (z, j) being the one whose
northeast vertex is (z, j). A permutation tt = X\X 2 - Xn is then represented
by putting an X in square (z,Xi), 1 < z < n. The next figure illustrates the
setup for the permutation
7t = 423651 7.
The reader should compare the following construction with Viennot’s in Sec-
tion 3.6.
X
X
X
X
X
X
X
We now define a growth : C^ that will depend on tt. We start by
letting
gAiJ) = 0 if i = 0 or j = 0.
Consider the square s with coordinates (z, j), labeled as in the diagram
i' P
s
A P
5.2. GROWTHS AND LOCAL RULES
199
Suppose, by induction on that we have defined — j —
and — 1, j), which we denote by A, /i, and i/, respectively. We then define
givihj)^ denoted by p, using the following local rules
LRl H jjL ^ u, then let p = p U i/.
LR2 If A ^ p = i/, then fi is obtained from A by adding 1 to A^ for some
i. Let p be obtained from p by adding 1 to Pi+i.
LR3 If A = p = z/, then let
_ J A if s does not contain an X,
^ [A with 1 added to Ai if s does contain an X.
Applying these rules to our example, we obtain the following diagram, where
for readability, parentheses and commas have been suppressed in the parti-
tions:
1
11
21
31
32
321
421
X
1
11
21
31
X
32
321
321
1
11
21
21
31
X
311
311
1
X
11
21
21
21
211
211
0
1
2
X
2
2
21
21
0
1
X
1
1
1
11
11
0
0
0
0
0
1
1
0 ^ 0 0 W 0 0 0 0 “
Our first order of business is to make sure that is well-defined.
200
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Lemma 5.2.2 Rules LRl-3 construct a well-defined growth : C^ Y in
that p is a partition and satisfies p ^ /x,
Proof. The only case where it is not immediate that p is a partition is LR2.
But by the way p is obtained from A we have pi > pi+i- So adding 1 to
will not disturb the fact that the parts are weakly decreasing.
It is clear that p covers or equals p,z/ except in LRl. Since X :<
by induction and p ^ there are three possibilities, namely X = p ■< u,
A = zy -< p, or A p, z/. It is easy to check that in each case, remembering
the proof of (2) in Proposition 5.1.3 for the third one, that p ^ p, z/. ■
The following lemma will also be useful. Its proof is a straightforward
application of induction and LRl-3, and so is left to the reader
Lemma 5.2.3 The growth has the following properties.
Y gTr{i,j — 1) ^ Qizihi) if only if there is an X in square (k,j) for
some k <i.
dirii — 1, j) ^ dniid) if only if there is an X in square (z,/) for
some I < j. ■
By this lemma, we have a saturated chain of partitions up the eastern
boundary of (7^,
0 = P7r(n,0) ^ g^{n, 1) < P7r(n,n),
corresponding to a standard tableau that we will denote by P^^. Similarly,
along the northern boundary we have defined by
0 = P7r(0,n) -< pvr(l,n) -< < P7r(n,n).
In our running example.
1357 1347
Pt^ = 2 6 and = 2 5
4 6
Comparing this with the tableaux obtained on page 94 (computed by insertion
and placement) gives credence to the following theorem.
Theorem 5.2.4 The map
IT y {Pm Qtt)
is the Robins on-Schens ted correspondence.
Proof. We will prove that P^^ = P{'^) since the equation Qtt = Q{'^) is
derived using similar reasoning.
First we will associate with each (i, j) in C^ a permutation iTij and a
tableau Pij. Note that in Pij^ the subscripts refer to the element of C^ and
not the entry of the tableau in row i and column j as is our normal convention.
5.2. GROWTHS AND LOCAL RULES
201
Since this notation will be used only in the current proof, no confusion will
result. Define iVij as the lower line of the partial permutation corresponding
to those X’s southwest of (i, j) in the diagram for g-j^. In our usual example,
7T6,3 = 231.
We also have a sequence of shapes
-^ 2,0 — ^ 2,1 ^
where Aj,/ = 0- So |Ai,//A^,/_i| = 0 or 1 for 1 < / < j. In the latter case,
label the box of with I to obtain the tableau Pij. These tableaux
are illustrated for our example in the following figure:
It is clear from the definitions that 7Tn,n = 'tt and Pn,n = Ptt- So it suffices
to prove that = P{'^ij) for all i,j < n. We will do this by induction on
i-\- j. It is obvious if z = 0 or j = 0. The proof breaks down into three cases
depending on the coordinates (z,/) of the X in row z, and {k,j) of the X in
column j.
202
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Case 1: I > j. (The case A: > z is similar.) By Lemma 5.2.3, we have
Xi j — l — so Pi^j — \ = Also, TTjjjf — i = SillCC Pi^j — \ — P(^7Ti^j — i^
by induction, we are done with this case.
Case 2: I = j. (Note that this forces k = i.) By Lemma 5.2.3 again,
Xi-ij-i = Xij-i = Xi-ij. So the second option in LR3 applies, and Pij
is Pi,j-i with j placed at the end of the first row. Furthermore, TTij is the
concatenation where j > max7Tij_i. So P{'7Tij) is also P{7Tij-i)
with j placed at the end of the first row, which finishes this case.
Case 3: I < j. (We can assume k < since the other possibilities are
taken care of by previous cases.) Using Lemma 5.2.3 one more time, we have
Xi-ij-i -< Xij-i, Xi-ij. Let c and d denote the cells of the skew partitions
Xij-i/Xi-ij-i and Xi-ij/Xi-ij-i, respectively. If c ^ d, then rule LRl is
the relevant one and Pi^j is Pi,j-\ with j placed in cell d. If c = d, then LR2
implies that Pij is Pij-i with j placed at the end of the next row below that
of d.
As far as the permutations are concerned, there are sequences cr, r such
that
— l — (Jt/,
~ ^ 3 '^'
where, as before, j > max7Tij_i. So the insertion paths for P(7Tij_i) and
P{7Tij) are the same except that when a path of the former ends at the row
currently occupied by j in the latter, then j is bumped to the next row. Now,
by induction, j ends up in cell d of P(7Ti_i^j), and the insertion path of I into
P(7r^_ij_i) ends in cell c of P{7Tij-i). Thus if c 7^ d, then j is not displaced
by the insertion of Z, and P{7Tij) is P{7Tij-i) with j placed in cell d. And if
c = d^ then j is displaced by ri into the row below d. For both options, the
description of P{7Tij) matches that of Pij^ so this completes the proof of the
theorem. ■
We are now in a position to define growths : C^ A where A is any
differential poset. The initial conditions should come as no surprise:
9n{hj) = 0 if i = 0 or j = 0.
We label the square s with coordinates (z, j) as follows, where a, 6, c are given
and d is to be defined:
c d
s
a b
To state the local rules, note that from condition DP2 we have, for fixed
|{a; : a; 6}| = \{y : y> 6}|.
5.2. GROWTHS AND LOCAL RULES
203
So for each b, pick a bijection
: {x : X <b} ^ {y : y>-b}.
The growth g,r will depend on the ^6, although our notation will not reflect
that fact.
DLRl If 6 7 ^ c, then a = b ^ c, a = c < b, ox a ^ b,c. In the first two
cases let d = max{6, c}. In the third let d be the unique element
covering b^c as guaranteed by DPS and Lemma 5.1.11.
DLR2 If a ^ 6 = c, then let d =
DLR3 If a = 6 = c, then let
^ _ f a if 5 does not contain an X,
~ I ^6(a) if 5 does contain an X.
It is easy to show that these rules are well-defined.
Lemma 5.2.5 Rules DLRl-3 construct a well-defined growth A
in that d satisfies d'^b^c. m
Furthermore, Lemma 5.2.3 continues to hold in this more general setting.
So, as before, we have saturated chains of partitions across the eastern and
northern boundaries of which we will denote by Ctt and respectively.
So Theorem 5.1.15 is a corollary of the following result.
Theorem 5.2.6 The map
7T — >
is a bijection between Sn and pairs of saturated 0-a chains as a varies over
all elements of rank n in the differential poset A.
Proof. Since both C and V end at (n, n) G it is clear that the corre-
sponding chains end at the same element which must be of rank n.
To show that this is a bijection, we construct an inverse. Given saturated
chains
C : 6 = Co ^ Cl ^ Cn
"D 0 = do di ^ dfi
where Cn = define
QTvip^j) = Cj and g^ii^Ti) = di for 0 < i, j < n.
Also, let Assuming that b,c^d are given around a square 5 as in
the previous diagram, we define a as follows.
RLDl If 6 ^ c, then d = b y c, d = c y b, ox d y b,c. In the first two
cases let a = min{fe, c}. In the third let a be the unique element
covered by b,c.
204
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
RLD2 U d y b = then let a = ^b{d)-
RLD3 If b = c = d, then let a = d.
Finally, we recover tt as corresponding to those squares where a = b = c ^
d. Showing that RLDl-3 give a well-defined growth associated with some
permutation tt and that this is the inverse procedure to the one given by
DLRl-3 is straightforward and left to the reader. ■
5.3 Groups Acting on Posets
The systematic study of group actions on posets was initiated by Stan-
ley [Stn 82] and has blossomed in recent years [Han 81, C-H-R 86, Sun 94,
B-W 95, S-Wa 97, S-We]. In this section and the next we will investigate
some of the interesting representations that arise in this context and some of
their combinatorial implications.
Definition 5.3.1 Let A be a finite, graded poset of rank n, having miminum
6 and maximum 1. Any S C {l,2,...,n — 1} defines a corresponding rank-
selected subposet
= {a G A : a = 6, a i, or rka G 5}
with partial order a < 6 in As if and only if the same inequality holds in A.
We will use the notation S' = {si, 52, . . . , 5fc}< to mean 5i < S 2 < • • • < 5/c. ■
For example, if A is the Boolean algebra B 4 and S — {1,3}, then the Hasse
diagram of As is given on the next page. All posets considered in this section
will be finite, graded, with a 0 and a 1, and so will admit rank-selections.
In order to define the invariants we wish to study, we will have to enlarge
our definition of representation.
Definition 5.3.2 Let group G have irreducible modules . . . ,
A virtual representation of G is a formal sum
V =
i
where the multiplicities are in Z, the integers. ■
If the rui are nonnegative, then V is an ordinary representation and the sum
can be interpreted as being direct.
Now let G be a group of automorphisms of the poset A, i.e., a group of
bijections g : A A such that both g and g~^ are order preserving. Then
it is easy to see that each g G G permutes the maximal (with respect to
containment) chains of As- Let A^ be the associated permutation module
for G. Define a virtual representation by
TCS
(5.6)
5.3. GROUPS ACTING ON POSETS
205
We should remark that Stanley [Stn 82] defined in terms of G’s action
on the reduced homology groups of As and then derived (5.6) as a corollary.
However, we will have no need here for that viewpoint.
We would also like to express A^ in term of the . This will be done
with the aid of the next result.
Theorem 5.3.3 Let V be a vector space and suppose we have two functions
f^g:Bn~^V. Then the following are equivalent:
(1) /(5) = Y. 9{T) for all S e Bn,
TCS
(2) 5(5) = ^(-l)l®-^l/(T)/or-an5e5„.
TCS
Proof. We will prove that (1) implies (2), as the converse is similar. Now,
^(_l)|S-T|/(j.) ^ 5](-l)l^-^l
TCS TCS UCT
UCS UCTCS
But from Exercise le in Chapter 4 we see that the inner sum is 0 unless
U = S. So the outer sum collapses to g{S). ■
206
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Theorem 5.3.3 is equivalent to the Principle of Inclusion-Exclusion (see
Exercise 17) which in turn is the special case of Mobius inversion (see Exer-
cise 18) where the underlying poset is the Boolean algebra [Sag 99]. As an
immediate corollary of this Theorem and equation (5.6), we have the following
result.
Corollary 5.3.4 For a// 5 C {1, 2, . . . , n - 1} we have
= ■
TCS
If the turn out to be ordinary representations, then something in-
teresting is probably going on combinatorially. This will be illustrated by
turning our attention to the Boolean algebra Bn- The symmetric group Sn
acts on the S' G in the natural way by
7r{ni,n2,...,n/,} = {Trni, 7rn2, . . . , Tm^}, (5.7)
and it is not hard to verify that this is an action, called an induced action,
which is an automorphism of Bn- In this case, if S = {ni, 722 , . . . , nfc}<, then
is equivalent to the representation afforded by the permutation module
M^, where M = (ni , 722 — ni , 723 — 722 , ... ,72 — Uk)- To see this, note that if
Ri^ R 2 ^ ' = {1, 2, ... , 72 } with = ni — ni-i for 1 < 2 < A; -h 1
(t2o = 0 and 72fc.fi = 72 ), then the map
~Ri
_jR2
C : 0 < Ri < R\ l±) R 2 < • • • < {1, 2, . . . , 72 } ^ (^*^)
extends linearly to an Sn -module isomorphism.
To describe B^, we need one more definition.
Definition 5.3.5 Let P be a standard tableau. Then 2 G P is a descent if
2 + 1 is in a lower row of P. The descent set of P is
DesP = {2 : 2 is a descent of P}. ■
For example,
12 3 5
if P = 4 6 8 , then DesP = {3, 5, 6}.
7
Counting descents is intimately connected with Kostka numbers.
Proposition 5 . 3.6 Let X \~ n, S = {721,722, . . . , 72 fc}< C {1, 2 , . . . , 72 - 1},
and /i = (t2i ,722 — 721 , . . . , 72 — 72 fc) . Then
|{P : Pa standard X-tableau and DesP C S}| = K\^.
5.3. GROUPS ACTING ON POSETS
207
Proof. We will give a bijection between the two sets involved, where
counts semistandard tableaux T of shape A and content jjl. Given P, replace
the numbers 1, 2, . . . , with ones, then ni -h 1, rii + 2, . . . , ri 2 with twos, and
so on. The next figure illustrates the map when A = (4,3, 1), S — {3,5,7},
and so = (3, 2, 2, 1). Below each T is the corresponding P, and below that,
Des P is given:
12 3 5
12 3 5
12 3 7
12 3 7
12 3 8
T:
4 6 7 ,
4 7 8 ,
4 5 6 ,
4 5 8 ,
4 5 7
8
6
8
6
6
1112
1112
1113
1113
1114
P:
2 3 3 ,
2 3 4 ,
2 2 3 ,
2 2 4 ,
2 2 3
4
3
4
3
3
DesP
{3,5,7},
{3,5},
{3,7},
{3,5,7},
{3,5}.
The resulting T is clearly a A-tableau of content /i, so we need only check
that it is semistandard. Since the rows of P increase and the replacement
map is weakly increasing, the rows of T must weakly increase. Similarly, the
columns of T weakly increase. To show that they actually increase strictly,
suppose Ti^j = Ti^ij = 1. Then at least one of ni-i + 1, n/_i + 2, . . . , n/ — 1
is a descent in P, contradicting Des PCS.
To show that this map is a bijection, we must construct an inverse. This
is a simple matter and left as an exercise. ■
We can now describe the decomposition of into irreducibles for the
Boolean algebra and show that in this case it is a genuine (as opposed to
virtual) representation.
Theorem 5.3.7 ([Stn 82]) For the action of Sn on Bn, decompose
A
Then the muliplicities are nonnegative integers, since
b^{\) = \{P : P a standard X-tableau and DesP — S}\.
Proof. Decompose as
A
Then Corollary 5.3.4 implies
aS(A) = ^ 6^(A) for all 5 C {1, 2, . . . ,n - 1}.
TCS
(5.9)
208
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
^
Define (respectively, b (A)) to be the number of standard A-tableaux P
with DesP C S (respectively, DesP = 5). Then directly from the definitions
for all S' C {1, 2, . . . ,n - 1}. (5.10)
TCS
Now, from Proposition 5.3.6 and the fact that is equivalent to we
have a^{X) = K\^ — a’^(A) for all S. But then inverting (5.9) and (5.10)
^
using Theorem 5.3.3 shows that b^{\) = b (A) for all 5, as desired. ■
5.4 Unimodality
Unimodality of a sequence of real numbers means that it rises to some max-
imum and then descends. It is the next most complicated behavior after
being weakly increasing or weakly decreasing. Unimodal sequences abound
in algebra, combinatorics, and geometry. See Stanley’s article [Stn 89] for a
survey. In this section we will show how group actions can be used to prove
unimodality.
Definition 5.4.1 A sequence {ak)k>Q = Uq, ui, U 2 , . . . of real numbers is uni-
modal if, for some index m.
^0 ^ ^1 ^ * ’ * — ^ ^ ' * * .
If the sequence is finite with n terms, then it is symmetric if
Ok = an-k for 0 < k < n.
A poset A with finite rank sets is symmetric or unimodal whenever the
sequence (\Ak\)k>o is. ■
Our first example of a symmetric, unimodal sequence comes from the
binomial coefficients.
Proposition 5.4*2 For any n, the sequence
is symmetric and unimodal. It follows that the same is true of the Boolean
algebra Bn .
Proof. For symmetry we have, from the factorial formula in Exercise lb of
Chapter 4,
n \ n\ n\ / n\
n — k) (n' — k)!(n — (n — k))! (n — k)!(k)! \k J
5.4. UNIMODALITY
209
Since the sequence is symmetric, we can prove unimodality by showing that
it increases up to its midpoint. Using the same exercise, we see that for
k < n/2:
(fc-i) ^ n!fc!(n - fc)! ^ k
n!(A: — l)!(n — A; + 1)! n — k 1
It will also be useful to talk about unimodal sequences of representations.
Definition 5.4.3 Let U, W be G-modules whose expansions in terms of the
G’s irreducibles are V = rriiV^'^^ and W = Yli>i Then we write
V <W if rrii < rii for all z > 1. ■
Using this partial order, we define a unimodal sequence of modules in the
obvious way.
To get permutation groups into the act, consider a finite graded poset, A.
Let Uk he the up operator of Definition 5.1.13 restricted to CA^. Also, let
X = Xk be the matrix of Uk in the bases {a : a E Ak} and {b : b E A^+i},
so
^ _ J 1 if a ^ 6,
~ \ 0 else.
Finally, let rkX denote the (linear- algebraic) rank of the matrix X.
Lemma 5.4.4 Let G he a group of automorphisms of a finite, graded poset
A.
1. IfrkX = \Ak\, then CAk < CA^+i.
2. IfrkX = \Ak-\-i\, then CAk > CAk+i.
Proof. We will prove the first statement, as the second is similar. We claim
that Uk is actually a G- homomorphism. It suffices to show that Ukigst.) =
gUkis.) for ^ e G and a E Ak. In other words, it suffices to prove that
{b : b y ga} = {gc : cy a}.
But if b >~ ga, then letting c = g~^b we have gc = b >- ga, so c V a. And if
cy a, then b = gey ga, so the two sets are equal.
Since rkA = \Ak\, we have that Uk is an injective G-homomorphism,
and so CAk is isomorphic to a submodule of CAk^i. Since G is completely
irreducible, the inequality follows. ■
Definition 5.4.5 A finite, graded poset A of rank n is ample if
rk Afc == min{|A/c|, |Afc+i|} for 0 < fc < n. ■
210
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
This is the definition needed to connect unimodality of posets and modules.
We will also be able to tie in the concept of orbit as defined in Exercise 2 of
Chapter 1. If G acts on a finite set S, then the orbit of s G 5 is
= { 9 ^ • 9 ^ ^}*
Also, let
S/G = {Os: se S}.
Theorem 5.4.6 ([Stn 82]) Let A he a finite, graded poset of rank n. Let G
be a group of automorphisms of A and V be an irreducible G -module. If A is
unimodal and ample, then the following sequences are unimodal.
(1) CAo,CAi,CA2,...,CA,.
(2) mo{V),mi{V),m 2 {V), . . . ,mn{V)y where rrik{V) is the multiplicity of
V in CAfc.
(3) |Ao/G|,|Ai/G|,|A2/G|,...,K/G|.
Proof. The fact that the first sequence is unimodal follows immediately from
the definition of ample and Lemma 5.4.4. This implies that the second se-
quence is as well by definition of the partial order on G-modules. Finally, by
Exercise 5b in Chapter 2, (3) is the special case of (2) where one takes V to
be the trivial module. ■
In order to apply this theorem to the Boolean algebra, we will need the
following lemma. The proof we give is based on Kantor’s*[Kan 72]. Another
is indicated in Exercise 25.
Proposition 5.4.7 The Boolean algebra Bn is ample.
Proof. By Proposition 5.4.2 and the fact that Xn-k-i is the transpose of
Xfc, it suffices to show that rkXk = (^) for k < n/2. Let Xs denote the
row ol X = Xk indexed by the set S. Suppose, towards a contradiction, that
there is a dependence relation
Xs=^J2asXs (5.11)
S^S
for certain real numbers as, not all zero. Let G be the subgroup of Sn that
fixes 5 as a set, i.e., G = S-g x where S is the complement of S in
{1, 2, . . . , n}. Then G acts on the fcth rank
Bn,k = {Bn)k,
and if ^ G G, then gS = S and Xgs,gT = Ns^t for all S G Bn,k, T G Bn^k+i-
Also, g permutes the S ^ S among themselves. So, using (5.11), we obtain
Xs,T ^ ^gS,gT = ^S,gT = °'SXs,gT = ^ agsXgS,gT = ^9sXs,T-
S^iS S^S SjiS
5.4. UNIMODALITY
211
It follows that and summing over all G G gives
m%=EE dgS^S = J2_XsJ2ags- (5.12)
9^G s^S S^S 9^G
By Exercise 5a in Chapter 2, the action of G partitions Bn,k into orbits
Oi. Directly from the definitions we see that we can write
Oi = {SeBn,k : |5n51 = 2}, 0<2<fc.
So Ok = {5} and is the set of all S ^ 5, 5 G Bn,k- Because of the
bijection from Exercise 2b in Chapter 1, if 5 G (9«, then as g runs over G,
the sets T = gS will run over Oi with each set being repeated |G|/|Oi| times.
So (5.12) can be rewritten
% =
for certain real numbers bi, not all zero (since the sum is a nonzero vector).
Now, for 0 < j < k, choose Tj G Bn,k+i with \Tj fl 5| = j, which can be done
since (A: + 1) + fc = 2A; + 1 < n. But then = 0. So the previous equation
for X^ gives
k-l
0 = E^*E^s,t,. (5.13)
i=0 SeOi
which can be viewed as a system of k equations in k unknowns, namely the
bi. Now, if 2 > j then Xs,Tj = 0 for all S £ 0{. So this system is actually
triangular. Furthermore, if i = j then there is an 5 G Oi with Xs^Tj = 1, so
that the system has nonzero coefficients along the diagonal. It follows that
bi = 0 for all 2 , the desired contradiction. ■
Corollary 5.4.8 Let G < Sn act on Bn and let V be an irreducible G-
module. Then, keeping the notation in the statement and proof of the previous
proposition, the following sequences are symmetric and unimodal.
(1) mo(F),mi(V'),m 2 (I^), . . . ,mn(V^); where mk{V) is the multiplicity of
V in CBn,fc*
(2) \Bn^o/G\, |^n,l/G|, \Bn,2/Gl . . . , \Bn,n/G\.
E E E
' ' 2=0 seOi geG
E E E
i=o seOi ' TeOi
E E E
2=0 '
E^>« E
2=0 SeOi
TeOi SeOi
212
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Proof. The fact that the sequences are unimodal follows from Theorem 5.4.6,
and Propositions 5.4.2 and 5.4.7. For symmetry, note that the map / : Bn,k —
Bn^n-k sending S to its complement induces a G-module isomorphism. ■
As an application of this corollary, fix nonnegative integers /c, I and con-
sider the partition (/^), which is just a fcx/ rectangle. There is a corresponding
lower order ideal in Young’s lattice
Yk,i = {\eY : A < (/*=)}
consisting of all partitions fitting inside the rectangle. We will show that this
poset is symmetric and unimodal. But first we need an appropriate group
action.
Definition 5.4.9 Let G and H be permutation groups acting on sets S and
T, respectively. The wreath product, G I H, acts on 5 x T as follows. The
elements ol Gl H are all pairs of the form (g, h), where h e H and g G
so ^ {gt : gt £ G). The action is
(g,h){s,t) = (gts,ht). (5.14)
It is easy to verify that G I H is group and that this is actually a group
action. ■
To illustrate, if G = <?3 and H = S 2 acting on { 1 , 2 , 3} and { 1 , 2 }, respectively,
then a typical element of Gl H is
( 9 ,h) = ig,{l,2)), where 5 = ( 51 ,^ 2 ) = ((1,2,3), (1,2,3)^).
So an example of the action would be
(5,h)(3,2) = (ff2(3),M2)) = (2,l).
Now, let
= ■ AC (/'=)}.
Corollary 5.4.10 For fixed k,l, the sequence
• • • ,PkAkl)
is symmetric and unimodal. So the poset Yk^i is as well
Proof. Identify the cells of (Z^) with the elements of { 1 , 2 , . . . , A:} x (1, 2, . . . , /}
in the usual way. Then SklSi has an induced action on subsets of (/^), which
are partially ordered as in Bki (by containment). There is a unique partition
in each orbit, and so the result follows from Corollary 5.4.8, part (2). ■
5 . 5 . CHROMATIC SYMMETRIC FUNCTIONS
213
5.5 Chromatic Symmetric Functions
In this section we will introduce the reader to combinatorial graphs and their
proper colorings. It turns out that one can construct a generating function
for such colorings that is symmetric and so can be studied in the light of
Chapter 4 .
Definition 5.5.1 A graphs F, consists of a finite set of vertices V = F(F)
and a set E" = E(F) of edges, which are unordered pairs of vertices. An edge
connecting vertices u and v will be denoted by uv. In this case we say that u
and V are neighbors. ■
As an example, the graph in the following figure has
V = {vi, V2, Vs, V4} and E = {ei, 62, 63, 64} == {viV2, V1V3, V2V3, V3V4},
so vi is is a neighbor of V2 and V3 but not of V4:
Definition 5.5.2 A coloring of a graph F from a color set C is a function
K :V -> C. The coloring is proper if it satifies
uv e E k{u) ^ n{v). ■
So a coloring just assigns a color to each vertex and, it is proper if no
two vertices of the same color are connected by an edge. One coloring of our
graph from the color set C = {1, 2, 3 } is
n{vi) = n{v2) = n{v4) = 1 and n{v3) — 2.
It is not proper, since both vertices of edge V\V2 have the same color. The
proper colorings of F are exactly those where , 1^2 5 ^3 all have distinct colors
and V4 has a color different from that of V3. Perhaps the most famous (and
controversial because of its computer proof by Appel and Haken) theorem
about proper colorings is the Four Color Theorem. It is equivalent to the fact
that one can always color a map with four colors so that adjacent countries
are colored differently.
214
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Theorem 5.5.3 (Four Color Theorem [A-H 76]) If graph T is planar
(i.e., can be drawn in the plane without edge crossings), then there is a proper
coloring k:V {1,2, 3, 4}. ■
Definition 5.5.4 The chromatic polynomial ofT is defined to be
Pp = Pi>{n) = the number of proper k : V {1,2, . . . ,n}. m
The chromatic polynomial provides an interesting tool for studying proper
colorings. Note that Theorem 5.5.3 can be restated: If T is planar, then
Fr(4) > 1. To compute -Pr(^) in onr running example, if we color the verticies
in order of increasing subscript, then there are n ways to color Vi. This
leaves n — 1 choices for V2, since viV2 G E, and n — 2 for vs. Finally, there
are n — 1 colors available for V4, namely any except n{vs). So in this case
Pr{n) = n{n — l)^(n — 2) is a polynomial in n. This is always true, as
suggested by the terminology, but is not obvious. The interested reader will
find two different proofs in Exercises 28 and 29.
In order to bring symmetric functions into play, one needs to generalize
the chromatic polynomial. This motivates the following definition of Stanley.
Definition 5.5.5 ([Stn 95]) If T has vertex set V = {^'l, . . . ,Vd}, then the
chromatic symmetric function of T is defined to be
~ A^r(x) ~ ^ ^ ^k{vi)^k{v2) ' ‘ * ^^{vd)^
where the sum is over all proper colorings k with colors from P, the positive
integers. ■
It will be convenient to let = ^k{vi)^k{v 2) ’ ’ '^K(vd)' return to our
example, we see that given any four colors we can assign them to V in 4\ = 24
ways. If we have three colors with one specified as being used twice, then it
is not hard to see that there are 4 ways to color T. Since these are the only
possibilities,
Ar(x) = 2AX1X2XSX4 + 24x1X2X3X5 H h 4xIx2Xs + 4xixlxs H
= 24mi4 + 4m2,i2.
We next collect a few elementary properties of Xr, including its connection
to the chromatic polynomial.
Proposition 5.5.6 ([Stn 95]) Let T be any graph with vertex set V .
1 . Xy is homogeneous of degree d = \V\.
2 . Xr is a symmetric function.
3. If we set X\ — = Xn = 1 and Xi =0 for i > n, written x = then
Xrin = PrinY
5.5. CHROMATIC SYMMETRIC FUNCTIONS
215
Proof. 1. Every monomial in Xr has a factor for each vertex.
2. Any permutation of the colors of a proper coloring gives another proper
coloring. This means that permuting the subscripts of Xr leaves the function
invariant. And since it is homogeneous of degree d, it is a finite sum of
monomial symmetric functions.
3. With the given specialization, each monomial of Xr becomes either 1
or 0. The surviving monomials are exactly those that use only the first n
colors. So their sum is Pr(^) by definition of the chromatic polynomial as
the number of such colorings. ■
Since Xr G A, we can consider its expansion in the various bases intro-
duced in Chapter 4. In order to do so, we must first talk about partitions of
a set S.
Definition 5.5.7 A set partition f3 = JB 1 /S 2 / • • • /^/ of S., (3 \~ 5, is a col-
lection of subsets, or blocks, Bi,. . . ,Bi whose disjoint union is 5. The type
of p is the integer partition
A(/3) = (|Si|,|B2|,...,|SH),
where we assume that the B{ are listed in weakly decreasing order of size. ■
For example, the partitions of {1, 2, 3, 4} with two blocks are
1,2, 3/4; 1,2, 4/3; 1,3, 4/2; 2, 3, 4/1; 1,2/3, 4; 1,3/2, 4; 1,4/2, 3.
The first four are of type (3, 1), while the last three are of type (2^).
Let us first talk about the expansion of Xr in terms of monomial sym-
metric functions.
Definition 5.5.8 An independent, or stable, set of T is a subset W C.V such
that there is no edge uv with both of its vertices in W. We call a partition
p\~V independent or stable if all of its blocks are. We let
'lx = = the number of independent partitions of V of type A. ■
Continuing with our example graph, the independent partitions of V are
Vilv2lv<^lv^', U1,U4/U2/U3; V2,v^/vi/v2, (5.15)
having types (T^) and (2, 1^). The fact that these are exactly the partitions
occuring in the previous expansion of Xr into monomial symmetric functions
is no accident. To state the actual result it will be convenient to associate
with a partition A = , . . . , the integer
y\ mi!m2! • -md\
Proposition 5.5.9 ([Stn 95]) The expansion of Xr in terms of monomial
symmetric functions is
Xr = 'Y^ixVxmx.
X\-d
216
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
Proof. In any proper coloring, the set of all vertices of a given color form
an independent set. So given k : V proper, the set of nonempty
2 G P, form an independent partition ph V. Thus the coefficient of in Xr
is just the number of ways to choose P of type A and then assign colors to
the blocks so as to give this monomial. There are i\ possiblities for the first
step. Also, colors can be permuted among the blocks of a fixed size without
changing x^, which gives the factor of y\ for the second. ■
To illustrate this result in our example, the list (5.15) of independent
partitions shows that
Xy = ^(i4)y(i4)m(i4) 4- i( 2 ,i 2 )^( 2 ,i 2 )m( 2 ,i 2 ) = 1 • 4!m(i4) 4- 2 • 2!m(2,i2),
which agrees with our previous calculations.
We now turn to the expansion of Xy in terms of the power sum sym-
metric functions. If F C E* is a set of edges, it will be convenient to let F
also stand for the subgraph of T with vertex set V and edge set F. Also,
by the components of a graph T we will mean the topologically connected
components. These components determine a partition P{F) of the vertex set,
whose type will be denoted by A(F). In our usual example, F — { 61 , 62 }
is a subgraph of T with two components. The corresponding partitions are
p{F) =vi,V 2 ,vs/v 4 and A(F) == (3,1).
Theorem 5.5.10 ([Stn 95]) We have
Xr= ^(-l)I^W(F).
FCE
Proof. Let K{F) denote the set of all colorings of T that are monochro-
matic on the components of F (which will usually not be proper). If P{F) —
B\j ... then we can compute the weight generating function for such
colorings as
^ X*" == 11(^1^'' + 4'®'' +•••)= Pa(f)(x).
k^K{F) 2=1
Now, for an arbitrary coloring k let E{f) be the set of edges uv of T such
that both k{u) = n{v). Directly from the definitions, k, G K{F) if and only if
E{k) D F. So
FCE
E E X-
FCE KeK{F)
E*“ E (-1)'^'
all K fce{k)
But from Exercise le in Chapter 4, the inner sum is 0 unless E{n) = 0, in
which case it is 1. Hence the only surviving terms are those corresponding to
proper colorings. ■
5.5. CHROMATIC SYMMETRIC FUNCTIONS
217
Returning to our running example, we can make the following chart, where
the last line is the number of F' C F with |F'| = |F| and A(F') = \{F)
(including F itself):
F
0
ei,C2
ei,C 4
ei, 62, 63
Cl, 62, 64
E
+1
-1
+1
+1
-1
-1
+1
\{F)
(1^)
(3,1)
~wv
(3,1)
(4)
(4)
F'
1
4
5
1
1
3
1
Plugging this information into Theorem 5.5.10 we get
= P(14) — 4p(2,l2) H- 5p(3^i) +P(2,2) ”P(3,1) “ 3p(4) +P(4)
= P(l4) - 4p(2,i2) +4p(3^1) +P(2,2) “ 2p(4).
The reader can check that this agrees with the earlier expansion in terms of
monomial symmetric functions.
Many other beautiful results about Xr have been derived [Stn 95, Stn ta,
S-S 93, Gas 96, G-S prl, G-S pr2], but there are also a number of open prob-
lems, one of which we would like to state here. A tree, T, is a connected
graph (only one component) that has no cycles, where a cycle is a sequence
of distinct vertices vi,V 2 , • • • ,Vn with viV 2 ,V 2 Vs, . . . ,Vn^vi G E{T).
Proposition 5.5.11 A tree T with \V\ = d has chromatic polynomial
Pt{ti) = n{n —
Proof. Pick any v eV that can be colored in n ways. Since T has no cycles,
each of the neighbors v can now be colored in — 1 ways. The same can be
said of the uncolored neighbors of the neighbors of v. Since T is connected,
we will eventually color every vertex this way, yielding the formula for Ft- ■
So all trees with d vertices have the same chromatic polynomial. But the
situation appears to be the opposite for Xt(x). To make this precise, we say
that graphs F and T are isomorphic if there is a bijection / : V{T) -> V{T),
called an isomorphism, such that uv € E{T) if and only if f{u)f{v) G E{T).
Question 5.5.12 ([Stn 95]) Let T and U be trees that are not isomorphic.
Is it true that
Xt{x) ^ Xu{x)7 m
This question has been answered in the affirmative up through \V\ = 9.
However, we should note that it is possible to find a pair of nonisomorphic
graphs that have the same chromatic symmetric function [Stn 95]. We hope
that the reader will be enticed to work on this problem and the many others
in the literature surrounding the symmetrie group.
218
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
5.6 Exercises
1. Prove that the poset of partitions under dominance is always a lattice,
but is not graded for n > 7.
2. Consider the The Fibonacci poset^ Z, defined as follows. The elements
of Z are all sequences w of Vs and 2’s with covering relation w ^ u if u
is obtained from w by either
Z1 deleting the first 1, or
Z2 picking some 2 preceded only by 2’s in w and changing it to a 1.
(a) Prove that Z is differential.
(b) Show that the number of elements of rank n is the nth Fibonacci
number, defined by Fq = = 1 and Fn — Fn-i + Fn -2 for n>2.
3. Consider the poset X of all rooted trees r (See Exercise 18 in Chapter 3)
ordered by inclusion.
(a) Show that X is a graded lattice .
(b) Show that the nth rank has Fn elements (see the previous exercise
for the definition of F^).
(c) Show that the saturated 0-r chains are in bijection with the nat-
ural labelings of r.
(d) Show that X is not differential, even though (by Exercise 18 in
Chapter 3) the satisfy a sum-of-squares formula.
4. Prove Proposition 5.1.14.
5. Prove Theorem 5.1.15.
6. Let r be a positive integer. A poset is r- differential if it satisfies Defi-
nition 5.1.10 with DP2 replaced by the following:
DP2r If a E A covers k elements for some k, then it is covered by k + r
elements.
Prove the following statements about r-differential posets A.
(a) The rank cardinalities \An\ are finite for all n > 0. (So the opera-
tors D and U are well-defined.)
(b) Let A be a graded poset with An finite for all n > 0. Then A is
r-differential if and only if DU — UD = rl.
(c) In any r-differential poset
'£{rf=r-n\,
CL^A-n
where is the number of saturated 0-a chains.
5.6. EXERCISES
219
(d) If A is r-differential and B is s-differential, then the product AxB
is (r + s)-differential. So if A is 1-difFerential, then the r-fold
product A'^ is r-differential.
7. Provide details for each of the three cases mentioned in Lemma 5.2.2.
8. In Theorem 5.2.4, fill in the proofs
(a) of the case k > i,
(b) that Qtt = Q(7t).
9. Come up with local rules for a growth Y such that the
corresponding Q'^ are the tableaux that would be those obtained by
using column insertion in the Robinson-Schensted algorithm.
10. Use the results of this chapter to prove that P{7t~^) = Q{n) and
g(7r-i) = P(7 t).
11. Show that LRl-3 are a special case of DLRl-3. What is the corre-
sponding function 4^^^?
12. (a) Prove Lemma 5.2.5.
(b) Show that any growth g : A with A differential must satisfy
DLRl.
13. Prove Lemma 5.2.3 for any g^^ : A with A differential.
14. Fill in the details of Theorem 5.2.6.
15. Show that if g is an automorphism of a poset A, then g permutes the
chains of any rank-selection As.
16. Prove (2) implies (1) in Theorem 5.3.3.
17. Let 5 be a set and let Si, S 2 , • • • , Sn be subsets of S. The set-theoretic
Principle of Inclusion- Exclusion states that
|S\ U 5,t = |5|- |5,|+ \SinSj\--- + {-ir\ fl 5^1
l<2<n l<i<n l<i<n
Prove this in two ways: using a direct argument and by applying The-
orem 5.3.3.
18. If A is a poset and a < 6 in A, then the corresponding (closed) interval
is
[a,b] = {c e A : a < c <b}.
Let IntA be the set of all closed intervals of A. Suppose that each
interval of A is finite. In this case we say that A is locally finite. The
220
CHAPTER 5, APPLICATIONS AND GENERALIZATIONS
Mobius function of A is the function /i : Int A -> Z defined recursively
by
{ -Ea<c<5M(a,c) if a <6,
where we write fi{a^h) for the more cumbersome /i([a,6]). We also use
yiA{cL^b) for /J^{a,b) if it is necessary to specify the poset.
(a) Show that (5.16) uniquely defines fi.
(b) Show that (5.16) is equivalent to
H{a,c) = 6a,b-
a<c<b
(c) Show that (5.16) is equivalent to
T, -= Sa,b-
a<c<b
(d) Posets A, A! are isomorphic if there is a bijection f : A A' such
that / and f~^ are both order preserving. Show that if A and A'
are isomorphic and /(a) = a', f{b) = b', then b) = /i^/(a', b').
(e) Prove that p respects poset products in that
fJ-AxA'i{a, a'), {b, b')) = ha{o,, b)nA'{a', b').
(f) Show that in the n-chain Cn we have
{ 1 if a = 6,
— 1 if a ^ 6,
0 otherwise.
(g) Show that Bn is isomorphic to the n-fold product Cf , and so
(h)
Prove the Mobius Inversion Theorem: Given a vector space F, a
locally finite poset A^ and two functions f,g:A-^V show that
/(®) = yz Va e A g{a) = ^ fi{b, a)f{b) Va £ A.
b<a b<a
and
/(a) = J29ib)
b>a
Ma £ A g{o) = ^ //(a, b)f{b) £ A
b>a
(i) Show that the Mobius Inversion Theorem implies Theorem 5.3.3.
5.6. EXERCISES
221
19. Verify that equation (5.7) defines an action of Sn that is an automor-
phism of Bn>
20. Show that equation (5.8) defines an 5n-isomorphism.
21. Construct an inverse for the map in Proposition 5.3.6.
22. Permutation tt = X\X 2 • • .Xn has a descent at index i\{xi> Xi^\. The
corresponding descent set is
DesTT — [i : z is a descent of tt}.
(a) Show that if, by Robinson-Schensted, Q{'k) = Q, then Deszr =
DesQ.
(b) Let X [- n, S — {ni, n 2 , . . . , nk}< C {1, 2, . . . , n — 1}, and jj, =
(ni,ri 2 - ni, . . . ,n - n/c). Then
|{7T G <Sn : Q(7 t) is a A-tableau and DesTr C 5}| =
(c) For the action of Sn on Bn, decompose
bS ^
A
Then
f^b^{X) = |{7T G Sn : Q(7t) is a A-tableau and DesTr = S}\.
23. Let (afc)fc>o = ao, ai, a 2 , . . . be a sequence of positive real numbers.
(a) Show that if the sequence of ratios {ak/cik-\-i)k>o is weakly increas-
ing, then the original sequence is unimodal.
(b) Show that if the original sequence is log concave (see Exercise 11
in Chapter 4), then it is unimodal.
24. Prove the second part of Lemma 5.4.4 in two ways: by mimicking the
proof of the first part and as a corollary to the statement of the first
part.
25. Give a different proof of Proposition 5.4.7 as follows. As before, it
suffices to show that vkXk = (^) for k < n/2. Let / be the (^) x (]J)
identity matrix.
(a) Show that
XkXi=Xl_,Xk-i + {n-k)I.
Hint; Use the definition of Xk to calculate the (5, T) entry on both
sides of the equation, where 5, T are subsets of {1, 2, . . . , n} with
k elements.
222
CHAPTER 5. APPLICATIONS AND GENERALIZATIONS
(b) Use the previous equation to show that XkXl is positive definite,
which implies that rk Xk = {^) •
26. If 0 < /c < / < n, let X = X{k, 1) be the matrix with rows (respectively,
columns) indexed by the A:-element (respectively, /-element) subsets of
{1,2,..., n} such that
1 if 5 C T,
0 otherwise.
Note that Xk = X{k,k P 1). Show that if /c -h / < n, then X{k, 1) has
full rank in two ways: by mimicking the proof of Proposition 5.4.7 and
by generalizing the previous exercise. For the latter you will need to
prove the identity
k
xik,i)x{k,iY
n — 2k
I — 2k “h i
X{i,kYX{i,k),
which can be demonstrated using Vandermonde’s convolution from Ex-
ercise If in Chapter 4.
27. Verify that G / FT is a group and that equation (5.14) does define an
action for the wreath product.
28. A theorem of Whitney states that
Pr(n) =
FCE
where 1{X{F)) is the length of A(F). Prove this using results from the
text. Note that as a corollary we have that Pr{n) is a polynomial in n.
29. (a) Given an edge e = uv e E'(P), let P \ e be the graph obtained
by deleting e from P. Also let P/e be the graph obtained by
contracting e to a point. So in P/e, u and v merge into a new
vertex x, and any edge wu or wv in P becomes an edge wx in P/e.
(If both wu and wv exist, then they merge into a single edge wx.)
Prove the Deletion- Contraction Law, which states that
Pr = Pr\e — Pr/e-
Use this recursion to prove the following facts about Pr by induc-
tion on the number of edges:
(b) Pr is a polynomial in n.
(c) degPr |1^|.
(d) The coefficients of Pr alternate in sign with leading coefficient 1.
(e) If P has k components, then the smallest power of n in Pr with
nonzero coefficient is n^.
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Index
Action, 7
on generalized tableaux, 79
on standard tableau.x, 55
on tabloids, 55
orbit, 48, 210
trivial, 11
Adjacent transposition, 4
Algebra, 4
Boolean, 58, 192
center of, 27
commutant, 23
endomorphism, 23
full matrix, 4
group, 8
Weyl, 197
Algorithm
backward slide, 113
deletion, 94
delta operator, 121
dual Knuth, 172
evacuation, 122
forward slide, 113
Greene-Nijenhuis-Wilf, 137
Hillman-Grassl, 148-150
insertion, 92
Knuth, 169-171
modified slide, 126-128
Novelli-Pak-Stoyanovskii, 125-
132
Robinson-Schensted, 92-94
straightening, 70
Alternant, 164
Ample, 209
Antidiagonal strip, 113
Appel-Haken, 213
Arm length of a hook, 125
Arm of a hook, 125
Automorphism, 204
Backward slide, 113
modified, 128
Ballot sequence, 176
Block, 215
Boolean algebra, 58, 192
Boundary of a partition, 117
Branching rule, 77
Candidate cell, 128
Cauchy identity, 171
Cell, 54
candidate, 128
Center, 27
of commutant algebra, 27, 29
of endomorphism algebra, 30
of matrix algebra, 27
Centralizer, 3
Chain
ascending, 194
descending, 194
saturated, 194
Character, 30
defining, 31
inner product, 34
linear, 31
orthogonality relations, 35
first kind, 35
second kind, 42
product, 168
regular, 31
table, 32
Characteristic map, 167, 168
Chromatic
polynomial, 214
symmetric function, 214
230
INDEX
231
Class function, 32
Code, 129
Coloring, 213
proper, 213
Column insertion, 95
Column tabloid, 72
Column-stabilizer, 60
Commutant algebra, 23
Complement, 13
orthogonal, 15
Complete homogeneous symmetric
functions, 152, 154
Complete reducibility, 17
Components, 216
Composition, 67
Conjugacy
class, 3
of group elements, 3
Content, 78
Converge, 144
Corner, 76
Cover, 58
Cycle, 2
in a graph, 217
of length fc, 2
type, 2
Cyclic module, 57
Decreasing subsequence, 97
Degree
of a monomial, 151
of a power series, 144
of a representation, 4
Deletion, 94
Delta operator, 121
Descent
permutation, 221
row, 70
tableau, 206
Determinantal formula, 132
Differential poset, see Poset, dif-
ferential
Dihedral group, 51
Direct sum, 13
Dominance lemma
for partitions, 59
for tableaux, 82
for tabloids, 68
Dominance order
for partitions, 58
for tableaux, 82
for tabloids, 68
Down operator
in a differential poset, 196
in Young’s lattice, 193
Dual Cauchy identity, 172
Dual equivalence of tableau, 117
Dual Knuth equivalence. 111
of tableaux, 113
Dual Knuth relations. 111
East, 129
Elementary symmetric functions, 152,
154
Endomorphism algebra, 23
Equivalence, 99
dual Knuth, 111
of tableaux, 113
dual tableau, 117
Knuth, 100
of tableaux, 113
P, 99
Q, 111
tableaux, 114
Equivalent
column tableaux, 72
modules, 19
row tableaux, 55
Euler, L., 144, 145
Evacuation, 122
Ferrers diagram, 54
Fibonacci
number, 218
poset, 218
Fixedpoint, 2
Fomin, S., 195, 197
Formal power series, 142
convergence of, 144
degree, 144
homogeneous of degree n, 151
Forward slide, 113
232
INDEX
modified, 126
Four Color Theorem, 214
Frame-Robinson-Thrall formula, 124
Frobenius reciprocity law, 48
Frobenius, G., 45, 166
Frobenius- Young formula, 132
Full matrix algebra, 4
Garnir element
of a pair of sets, 70
of a tableau, 71
General linear group
of a vector space, 6
of matrices, 4
Generalized permutation, 169
Generalized Young tableau, 78
Generating function, 142
weight, 145
Gessel-Viennot, 158
Graded, 195
Grading, 152
Graph, 213
chromatic polynomial, 214
coloring, 213
components, 216
cycle, 217
Four Color Theorem, 214
independent partition, 215
independent set, 215
isomorphism, 217
neighbors, 213
proper coloring, 213
stable partition, 215
stable set, 215
tree, 217
Greatest lower bound, 192
Greene, C., 102, 173
Greene-Nijenhuis-Wilf algorithm, 137
Group
action, see Action
algebra, 8
character, 30
cyclic, 5
dihedral, 51
general linear, 4, 6
symmetric, 1
Growth, 197
Haiman, M. D., 117
Hasse diagram, 58
normal, 112
skew, 112
Hillman-Grassl algorithm, 148-150
Homogeneous of degree n, 151
Homomorphism, 18
corresponding to a tableau, 80
Hook, 124
arm, 125
formula, 124
leg, 125
rim or skew, 180
rooted tree, 138
shifted, 139
tableau, 126
Hooklength, 124
rooted tree, 138
shifted, 139
Inclusion-Exclusion, see Principle
of Inclusion-Exclusion
Incomparable, 58
Increasing subsequence, 97
Independent
partition, 215
set, 215
Induced representation, 45
Inequivalent, 19
Inner corner, 76
Inner product
invariant, 14
of characters, 34
Insertion
column, 95
path, 92
row, 92
Insertion tableau, 93
Interval of a poset, 219
Invariant
inner product, 14
subspace, 10
Involution, 2
Irreducibility, 12
INDEX
233
Isomorphism
graph, 217
module, 19
poset, 220
Jacobi- Trudi determinants, 158, 182
James, G. D., viii, 53, 63, 66, 182
Jeu de taquin, 116, 173
backward slide, 113
forward slide, 113
modified backward slide, 128
modified forward slide, 126
Join, 192
/c-decreasing subsequence, 103
fc-increasing subsequence, 103
Knuth, D. E.
algorithm, 169-171
dual algorithm, 172
dual equivalence. 111
of tableaux, 113
dual relations. 111
first kind. 111
second kind. 111
equivalence, 100
of tableaux, 113
relations, 99, 173
Kostka numbers, 85
Lattice
path, 158
involution, 161
labelings, 158-159
sign, 160
weight, 159-160
poset, 192
Young, 192
down operator, 193
up operator, 193
Lattice permutation, 176
Least upper bound, 192
Ledermann, W., vii
Leg length
of a hook, 125
of a rim or skew hook, 180
Leg of a hook, 125
Lemma
dominance, see Dominance lemma
Schur’s, 22
sign, 64
Length, 152
Length of a subsequence, 98
Lexicographic order, 59
Littlewood-Richardson
coefficient, 175
rule, 177, 183
Local rules, 199
Locally finite, 219
Log concave, 189
Lower order ideal, 85
Mobius
function, 219
inversion, 206, 219
inversion theorem, 220
Macdonald, I. G., viii, 164
Maschke’s theorem
for matrices, 17
for modules, 16
Matrix, 4
direct sum, 13
tensor product, 25
Maximal element, 69
Maximum element, 69
Meet, 192
Miniature tableau, 118
Minimal element, 69
Minimum element, 69
Modified slide, 126-128
Module, 6
complement, 13
cyclic, 57
direct sum, 13
endomorphism algebra, 23
equivalent, 19
generated by a vector, 57
homomorphism, 18
inequivalent, 19
irreducible, 12
isomorphism, 19
permutation module M^, 56
reducible, 12
234
INDEX
Specht, 62
tensor product, 44
unimodal sequence, 209
Monomial symmetric function, 151
Multiplicative, 154
Murnaghan-Nakayama rule, 180
n-chain, 58, 192
Natural labeling, 150
Neighbors, 213
Normal diagram, 112
North, 129
step, 129, 158
Novelli-Pak-Stoyanovskii algorithm,
125-132
Orbit, 48, 210
Order
dominance, see Dominance or-
der
lexicographic, 59
preserving, 197
Orthogonal complement, 15
Outer corner, 76
P-equivalence, 99
P-tableau, 93
Partial order, 58
Partial permutation, 101
Partial tableau, 92
Partially ordered set, see Poset
Partition, 2
boundary, 117
dominance
lemma, 59
order, 58
hook, 124
hooklength, 124
length, 152
normal, 112
set, 215
skew, 112
strict, 139
Path
Gessel-Viennot, 158
Hillman-Grassl, 148
insertion, 92
Novelli-Pak-Stoyanovskii, 127
reverse Hillman-Grassl, 149
reverse Novelli-Pak-Stoyanovskii,
128
Permutation, 1
adjacent transposition, 4
cycle, see Cycle
fixedpoint, 2
generalized, 169
involution, 2
lattice, 176
matrix, 6
module M^, 56
one-line notation, 2
partial, 101
representation, 8
reversal, 97
shadow diagram, 108
shadow line, 107
skeleton of, 110
transposition, 4
adjacent, 4
two-line notation, 1
Permutation matrix, 6
Placement, 93
Polytabloid, 61
Poset, 58
ample, 209
ascending chain, 194
automorphism, 204
Boolean algebra, 58, 192
chain, 194
covering relation, 58
descending chain, 194
differential, 196
down operator, 196
Fibonacci, 218
r-differential, 218
up operator, 196
graded, 195
greatest lower bound, 192
growth, 197
Hasse diagram, 58
incomparable elements, 58
interval, 219
INDEX
235
isomorphism, 220
join, 192
lattice, 192
least upper bound, 192
locally finite, 219
lower order ideal, 85
maximal element, 69
maximum element, 69
meet, 192
minimal element, 69
minimum element, 69
n-chain, 58, 192
natural labeling, 150
order preserving, 197
product, 197
r-differential, 218
rank, 195
rank-selected, 204
refinement of, 59
reverse partition, 150
saturated chain, 194
symmetric, 208
unimodal, 208
Power sum symmetric functions, 152,
154
Principle of Inclusion-Exclusion, 206,
219
Product of characters, 168
Product of posets, 197
Proper coloring, 213
Q-equi valence, 111
Q-tableau, 93
r-differential poset, 218
Rank, 195
Rank-selected poset, 204
Reciprocity law of Probenius, 48
Recording tableau, 93
Reducibility, 12
complete, 17
Refinement, 59
Representation, 4
completely reducible, 17
coset, 9, 20
defining, 5, 8, 11, 14, 18, 20
degree, 4
direct sum, 13
induced, 45
irreducible, 12
matrix, 4
of cyclic groups, 5
permutation, 8
reducible, 12
regular, 8, 11
restricted, 45
sign, 5, 11, 19
tensor product, 43
trivial, 4, 11, 18
Young’s natural, 74
Restricted representation, 45
Reversal, 97
Reverse ballot sequence or lattice
permutation, 176
Reverse path
code, 129
Hillman-Grassl, 149
Novelli-Pak-Stoyanovskii, 128
Reverse plane partition, 147
Rim hook, 180
tableau, 184
Robinson- Schensted algorithm, 92-
94
Roby, T. W., 197
Rooted tree, 138
hook, 138
hooklength, 138
Row word, 101
Row-stabilizer, 60
Rule
branching, 77
Littlewood- Richardson, 177, 183
local, 199
Murnaghan-Nakayama, 180
Young, 85, 174
Saturated, 194
Schiitzenberger, M.-P., 106, 112, 121,
173, 177
Schur function, 155
skew, 175
Schur’s lemma
236
INDEX
for matrices, 22
for modules, 22
over complex field, 23
Semistandard
basis, 82
Young tableau, 81
Sequence
symmetric, 208
unimodal, 208
Set partition, 215
block, 215
independent, 215
stable, 215
type, 215
Shadow, 107
diagram, 108
line, 107
x-coordinate, 108
^/-coordinate, 108
Shape, 54
rooted tree, 138
shifted, 139
skew, 112
Shifted
hook, 139
hooklength, 139
shape, 139
Sign
of a permutation, 4
of a rim hook tableau, 185
of lattice paths, 160
representation, 5, 11, 19
Sign lemma, 64
Skeleton, 110
Skew
diagram, 112
hook, 180
Skew-symmetric functions, 163
Slide, 113
modified backward, 128
modified forward, 126
sequence, 114
South, 129
Specht module, 62
Stable
partition, 215
set, 215
Standard
basis, 67
Young tableau, 66
Stanley, R. P., 150, 195-197, 214
Step
eastward, 158
northward, 129, 158
westward, 129
Straightening algorithm, 70
Strict partition, 139
Submodule, 10
theorem, 65
Subsequence, 97
decreasing, 97
increasing, 97
/^-decreasing, 103
/^-increasing, 103
length, 98
Superstandard tableau, 157
Symmetric
group, 1
poset, 208
sequence, 208
Symmetric function
chromatic, 214
complete homogeneous, 152, 154
elementary, 152, 154
grading, 152
monomial, 151
multiplicative, 154
power sum, 152, 154
ring of, 151
Schur, 155
skew, 175
Tableau, 55
antidiagonal strip, 113
column-stabilizer, 60
content, 78
delta operator, 121
descent, 206
dominance
lemma, 82
order, 82
dual equivalence, 117
INDEX
237
equivalent, 114
evacuation, 122
Garnir element, 71
generalized, 78
hook, 126
insertion tableau, 93
miniature, 118
of shape A, 55
P-tableau, 93
partial, 92
Q-tableau, 93
recording tableau, 93
rim hook, 184
row descent, 70
row equivalent, 55
row word, 101
row-stabilizer, 60
semistandard, 81
standard, 66
superstandard, 157
type, 78
union, 120
Tabloid, 55
column, 72
dominance
lemma, 68
order, 68
of shape A, 55
Tensor product
of matrices, 25
of representations, 43
of vector spaces, 26
Theorem
Mobius inversion, 220
Maschke
for matrices, 17
for modules, 16
submodule, 65
Thomas, G. P., 177
Total order, 58
Totally ordered set, 58
Transposition, 4
adjacent, 4
Tree, 217
Type
cycle, 2
set partition, 215
tableau, 78
Unimodal
module sequence, 209
poset, 208
sequence, 208
Union of tableaux, 120
Up operator
in a differential poset, 196
in Young’s lattice, 193
Vandermonde
convolution, 186
determinant, 164
Vector space, 6
direct sum, 13
orthogonal complement, 15
tensor product, 26
Viennot, G., 106
Weakly
east, 129
north, 129
south, 129
west, 129
Weight, 145
generating function, 145
of lattice paths, 159-160
tableau, 155
West, 129
step, 129, 158
Weyl algebra, 197
Wreath product, 212
Young, 54
lattice, 192
down operator, 193
up operator, 193
natural representation, 74
polytabloid, 61
rule, 85, 174
subgroup, 54
tableau
generalized, 78
of shape A, 55
semistandard, 81
238
INDEX
standard, 66
tabloid, 20
of shape A, 55
Graduate Texts in Mathematics
(continued from page ii)
66 Waterhouse. Introduction to Affine
Group Schemes.
67 Serre. Local Fields.
68 Weidmann. Linear Operators in Hilbert
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69 Lang. Cyclotomic Fields II.
70 Massey. Singular Homology Theory.
71 Farkas/Kra. Riemann Surfaces. 2nd ed.
72 Stillwell. Classical Topology and
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73 Hungerford. Algebra.
74 Davenport. Multiplicative Number
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75 Hochschild. Basic Theory of Algebraic
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76 IiTAKA. Algebraic Geometry.
77 Hecke. Lectures on the Theory of
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78 Burris/S ANKAPPANAVAR. A Course in
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79 Walters. An Introduction to Ergodic
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80 Robinson. A Course in the Theory of
Groups. 2nd ed.
8 1 Forster. Lectures on Riemann Surfaces.
82 Bott/Tu. Differential Forms in Algebraic
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83 Washington. Introduction to Cyclotomic
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84 Ireland/Rosen. A Classical Introduction
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85 Edwards. Fourier Series. Vol. II. 2nd ed.
86 VAN Lint. Introduction to Coding Theory.
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87 Brown. Cohomology of Groups.
88 Pierce. Associative Algebras.
89 Lang. Introduction to Algebraic and
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90 Brondsted. An Introduction to Convex
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9 1 Beardon. On the Geometry of Discrete
Groups,
92 Diestel. Sequences and Series in Banach
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93 Dubrovin/Fomenko/Novikov. Modem
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94 Warner. Foundations of Differentiable
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95 Shiryaev. Probability. 2nd ed.
96 Conway. A Course in Functional
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97 Koblitz. Introduction to Elliptic Curves
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98 Brocker/Tom Dieck. Representations of
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99 Grove/Benson. Finite Reflection Groups.
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100 Berg/Christensen/Ressel. Harmonic
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101 Edwards. Galois Theory.
102 Varadaraj AN. Lie Groups, Lie Algebras
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103 Lang. Complex Analysis. 3rd ed.
104 Dubrovin/Fomenko/Novikov. Modem
Geometry — Methods and Applications.
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105 Lang. 5'l2(R).
106 Silverman. The Arithmetic of Elliptic
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107 Olver. Applications of Lie Groups to
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108 Range. Holomorphic Functions and
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1 09 Lehto. Univalent Functions and
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1 10 Lang. Algebraic Number Theory.
1 1 1 Husemoller. Elliptic Curves.
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117 Serre. Algebraic Groups and Class Fields.
1 1 8 Pedersen. Analysis Now.
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120 ZiEMER. Weakly Differentiable Functions:
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121 Lang. Cyclotomic Fields I and II.
Combined 2nd ed.
122 Remmert. Theory of Complex Functions.
Readings in Mathematics
123 Ebbinghaus/Hermes et al. Numbers.
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124 Dubrovin/Fomenko/Novikov. Modem
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125 Berenstein/Gay. Complex Variables: An
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126 Borel. Linear Algebraic Groups. 2nd ed.
127 Massey. A Basic Course in Algebraic
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128 Rauch. Partial Differential Equations.
129 Fulton/Harris. Representation Theory: A
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130 Dodson/Poston. Tensor Geometry.
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132 Beardon. Iteration of Rational Functions.
133 Harris. Algebraic Geometry: A First
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134 Roman. Coding and Information Theory.
135 Roman. Advanced Linear Algebra.
136 Adkins/Weintraub. Algebra: An
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137 Axler/Bourdon/Ramey. Harmonic
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138 Cohen. A Course in Computational
Algebraic Number Theory.
139 Brecon. Topology and Geometry.
140 Aubin. Optima and Equilibria. An
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141 Becker/ Weispfenning/Kredel. Grobner
Bases. A Computational Approach to
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142 Lang. Real and Functional Analysis.
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143 Doob. Measure Theory.
144 Dennis/Farb. Noncommutative
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145 Vick. Homology Theory. An
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2nd ed.
146 Bridges. Computability: A
Mathematical Sketchbook.
147 Rosenberg. Algebraic A:-Theory
and Its Applications.
148 ROTMAN. An Introduction to the
Theory of Groups. 4th ed.
149 Ratcliffe. Foundations of
Hyperbolic Manifolds.
150 Eisenbud. Commutative Algebra
with a View Toward Algebraic
Geometry.
1 5 1 Silverman. Advanced Topics in
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152 Ziegler. Lectures on Polytopes.
153 Fulton. Algebraic Topology: A
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154 Brown/Pearcy. An Introduction to
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155 Kassel. Quantum Groups.
156 Kechris. Classical Descriptive Set
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157 M ALLI A VIN. Integration and
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158 Roman. Field Theory.
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160 Lang. Differential and Riemannian
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162 Alperin/Bell. Groups and
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163 Dixon/Mortimer. Permutation Groups.
164 Nathanson. Additive Number Theory:
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165 Nathanson. Additive Number Theory:
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166 Sharpe. Differential Geometry: Cartan's
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167 Morandi. Field and Galois Theory.
168 Ewald. Combinatorial Convexity and
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169 Bhatia. Matrix Analysis.
170 Bredon. Sheaf Theory. 2nd ed.
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1 72 Remmert. Classical Topics in Complex
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173 Diestel. Graph Theory. 2nd ed.
174 Bridges. Foundations of Real and
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175 Lickorish. An Introduction to Knot
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176 Lee. Riemannian Manifolds.
177 Newman. Analytic Number Theory.
1 78 Clarke/Led yaev/Stern/Wolenski.
Nonsmooth Analysis and Control
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179 Douglas. Banach Algebra Techniques in
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180 Srivastava. a Course on Borel Sets.
1 8 1 Kress. Numerical Analysis.
1 82 Walter. Ordinary Differential
Equations.
1 83 Megginson. An Introduction to Banach
Space Theory.
184 Bollobas. Modem Graph Theory.
185 Cox/Little/O'Shea. Using Algebraic
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186 Ramakrishnan/Valenza. Fourier
Analysis on Number Fields.
1 87 Harris/Morrison. Moduli of Curves.
1 88 Goldblatt. Lectures on the Hyperreals:
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189 Lam. Lectures on Modules and Rings.
190 Esmonde/Murty. Problems in Algebraic
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1 9 1 Lang. Fundamentals of Differential
Geometry.
192 Hirsch/Lacombe. Elements of Functional
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193 Cohen. Advanced Topics in
Computational Number Theory.
194 Engel/Nagel. One-Parameter Semigroups
for Linear Evolution Equations.
195 Nathanson. Elementary Methods in
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196 Osborne. Basic Homological Algebra.
197 Eisenbud/Harris. The Geometry of
Schemes.
198 Robert. A Course in /?-adic Analysis.
199 Hedenmalm/Korenblum/Zhu. Theory
of Bergman Spaces.
200 Bao/Chern/Shen. An Introduction to
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201 Hindry/Silverman. Diophantine
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202 Lee. Introduction to Topological
Manifolds.
203 Sagan. The Symmetric Group:
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2nd ed.
204 Escofier. Galois Theory.
205 Felix/Halperin/Thomas. Rational
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206 Murty. Problems in Analytic Number
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207 Godsil/Royle. Algebraic Graph Theory.
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